Chapter 13.14 - Graph of Derivatives

In the previous section, we saw the general form of the derivative. We saw some solved examples also. In this section, we will see a few more solved examples. We will also see the graphs of derivatives.

Solved example 13.10
Find the derivative of f(x) = 10x.
Solution:
• In our present case, f(x) = 10x. So we get:
$\begin{array}{ll}
{}&{f'(x)}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{f(x+h) – f(x)}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{10(x+h) – 10 x}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{10x + 10h – 10x}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{10h}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {10 \lim_{h\rightarrow 0}{\left[\frac{h}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {10 \lim_{h\rightarrow 0}{1}}
&{} \\

{}&{}
& {~=~}& {10 × 1}
&{} \\

{}&{}
& {~=~}& {10}
&{} \\

\end{array}$

◼ A graphical description can be written in 5 steps:
1. In the previous section, we saw that, f’(x) is a function. So we must  be able to plot f’(x).
2. In fig.13.34 below, both f(x) and f’(x) are plotted in the same graph.

Fig.13.34

• In order to accommodate both functions in the same graph, the scale is changed to the following values:
    ♦ x-axis: 1 unit = 2
    ♦ y-axis: 1 unit = 10
3. The two graphs are plotted in different colors.
    ♦ The red line represents f(x)
    ♦ The green line represents f’(x)
4. f’(x) is a constant function. So, as expected, the green line is horizontal.
5. Let us see the practical application of the graph. It can be written using two examples:
(i) Suppose that, we want the derivative of f(x) at (x = 1.5). That is., we want f’(1.5)
• For that, we mark point A on the red line. Here, (x = 1.5)
• Next we draw a vertical line through A. This vertical line meets the green line at A'.
• f’(1.5) will be equal to the y-coordinate of A'.
(ii) Suppose that, we want the derivative of f(x) at (x = -2.3). That is., we want f’(-2.3)
• For that, we mark point B on the red line. Here, (x = -2.3)
• Next we draw a vertical line through B. This vertical line meets the green line at B'.
• f’(-2.3) will be equal to the y-coordinate of B'.

Solved example 13.11
Find the derivative of f(x) = x² .
Solution:
• In our present case, f(x) = x². So we get:
$\begin{array}{ll}
{}&{f'(x)}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{f(x+h) – f(x)}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{[(x+h)^2] – [x^2]}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{[(x^2 + 2hx + h^2)] – [x^2]}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{2hx + h^2}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{h(h +2x)}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{[h + 2x]}}
&{} \\

{}&{}
& {~=~}& {2x}
&{} \\

\end{array}$

◼ A graphical description can be written in 5 steps:
1. In the previous section, we saw that, f’(x) is a function. So we must  be able to plot f’(x).
2. In fig.13.35 below, both f(x) and f’(x) are plotted in the same graph.

Fig.13.35

3. The two graphs are plotted in different colors.
    ♦ The red curve represents f(x)
    ♦ The green line represents f’(x)
4. Let us see the practical application of the graph. It can be written using two examples:
(i) Suppose that, we want the derivative of f(x) at (x = 3). That is., we want f’(3)
• For that, we mark point A on the red line. Here, (x = 3)
• Next we draw a vertical line through A. This vertical line meets the green line at A'.
• f’(3) will be equal to the y-coordinate of A'. In this case, it is 6.
• We can verify this theoretically:
f`(3) = 2 × 3 = 6
(ii) Suppose that, we want the derivative of f(x) at (x = -2.5). That is., we want f’(-2.5)
• For that, we mark point B on the red line. Here, (x = -2.5)
• Next we draw a vertical line through B. This vertical line meets the green line at B'.
• f’(-2.5) will be equal to the y-coordinate of B'. In this case, it is -5.
• We can verify this theoretically:
f`(-2.5) = 2 × -2.5 = -5
5. Further more, the reader may verify the tangents also:
• Through the point A, draw a line at a slope of 6. This line will be tangential to f(x).
• Through the point B, draw a line at a slope of -5. This line will be tangential to f(x). 

Solved example 13.12
Find the derivative of the constant function f(x) = a for a fixed real number a.
Solution:
• In our present case, f(x) = a. So we get:

$\begin{array}{ll}
{}&{f'(x)}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{f(x+h) – f(x)}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{[a] – [a]}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{0}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {0}
&{} \\

\end{array}$

Solved example 13.13
Find the derivative of f(x) = $\frac{1}{x}$.
Solution:
• In our present case, f(x) = $\frac{1}{x}$. So we get:
$\begin{array}{ll}
{}&{f'(x)}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{f(x+h) – f(x)}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\left[\frac{1}{x+h}\right] – \left[\frac{1}{x}\right]}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\frac{x - x - h}{x(x+h)}}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{-h}{xh(x+h)} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{-1}{x(x+h)} \right]}}
&{} \\

{}&{}
& {~=~}& {\frac{-1}{x(x+0)}}
&{} \\

{}&{}
& {~=~}& {-\frac{1}{x^2}}
&{} \\

\end{array}$

◼ A graphical description can be written in 5 steps:
1. In the previous section, we saw that, f’(x) is a function. So we must  be able to plot f’(x).
2. In fig.13.36 below, both f(x) and f’(x) are plotted in the same graph.

Fig.13.36

3. The two graphs are plotted in different colors.
    ♦ The red curve represents f(x)
    ♦ The green curve represents f’(x)
4. Let us see the practical application of the graph. It can be written using two examples:
(i) Suppose that, we want the derivative of f(x) at (x = 1.5). That is., we want f’(1.5)
• For that, we mark point A on the red line. Here, (x = 1.5)
• Next we draw a vertical line through A. This vertical line meets the green line at A'.
• f’(1.5) will be equal to the y-coordinate of A'. In this case, it is -0.44.
• We can verify this theoretically:
f`(1.5) = $-\frac{1}{1.5^2}$ = -0.44
(ii) Suppose that, we want the derivative of f(x) at (x = -2). That is., we want f’(-2)
• For that, we mark point B on the red line. Here, (x = -2)
• Next we draw a vertical line through B. This vertical line meets the green line at B'.
• f’(-2) will be equal to the y-coordinate of B'. In this case, it is -0.25.
• We can verify this theoretically:
f`(-2) = $-\frac{1}{(-2)^2}$ = -0.25
5. Further more, the reader may verify the tangents also:
• Through the point A, draw a line at a slope of 0.-44. This line will be tangential to f(x).
• Through the point B, draw a line at a slope of -0.25. This line will be tangential to f(x).

---

In the next section, we will see algebra of derivative of functions.

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Chapter 13.13 - Derivative as a Function

In the previous section, we saw an easy method to find the derivative at a point. In this section, we will see a general form.

• The general form can be obtained using the same fig.13.30 that we saw in the previous section. All we need to do is, change 'a' to 'x'.
• The steps are the same with 'x' in place of 'a'. However, we will write all those 11 steps here:
1. In fig.13.31 below, the red curve is the graph of f(x).

Fig.13.31

• Two points P and Q are marked on the red curve.
(P is an arbitrary point. An arbitrary point is chosen so that, it will be applicable for all cases. The dictionary meaning of the word "arbitrary" can be seen here) 
    ♦ A green horizontal line is drawn through P.
    ♦ A green vertical line is drawn through Q.
    ♦ These horizontal and vertical lines meet at R
• Thus we get a right triangle PQR
2. A green dashed vertical line is drawn through P.
• This vertical line meets the x-axis at (x,0)
[We are able to write "(x,0)" because, P is an arbitrary point. Since it is an arbitrary point, 'x' can be any point in the domain of f(x)]
• So we can write:
The x-coordinate of P is ‘x’.
• If the x-coordinate is ‘x’, then obviously, the y-coordinate will be f(x)
• Thus we get the coordinates of P: (x,f(x))
3. Another green dashed vertical line is drawn through R.
• This vertical line meets the x-axis at ((x+h),0)
• So we can write:
The x-coordinate of Q is ‘(x+h)’.
• If the x-coordinate is ‘(x+h)’, then obviously, the y-coordinate will be f(x+h)
• Thus we get the coordinates of Q: ((x+h),f(x+h))
4. Using the x-coordinates of P and Q, we can find the horizontal distance between P and Q.
• We get:
Horizontal distance between P and Q = [(x+h) – x] = h
• That means, the length PR = h.
5.  Using the y-coordinates of P and Q, we can find the vertical distance between P and Q.
• We get:
Vertical distance between P and Q = [f(x+h) – f(x)]
• That means, the length QR = [f(x+h) – f(x)].
6. Now we have the base and altitude of the right triangle PQR.
• So we can write the slope of the line PQ.
Slope of PQ = $\frac{[f(x+h) – f(x)]}{h}$
7. If Q is brought very close to P, then the slope calculated in (6) will be the slope of the tangent at P
• For bringing Q closer to P, we must decrease the length h. When h approaches zero, Q will be very close to P.
• h must become very close to zero. At the same time, it must not become exact zero. This can be written as:
$\lim_{h\rightarrow 0} h$
8. From (5), we have the length of the altitude:
[f(x+h) – f(x)]
• When h approaches zero, the base PR will become infinitesimal.
• There will be corresponding changes in the altitude also.
• When h approaches zero, the length of the altitude can be written as: $\lim_{h\rightarrow 0} {[f(x+h) – f(x)]}$
9. So now we can write the slope of the tangent at P:
$\frac{\lim_{h\rightarrow 0} {[f(x+h) – f(x)]}}{\lim_{h\rightarrow 0} h}$
• The limit is being applied to both numerator and denominator. So this can be written in a simplified form as:
Slope of tangent at P = $\lim_{h\rightarrow 0}{\left[\frac{f(x+h) – f(x)}{h} \right]}$
10. But slope of the tangent at P is the derivative at P.
• Point P corresponds to any x in the domain.
• So we can write:
The result in (9) gives the derivative of f(x) at any x in the domain.
11. The derivative of f(x) at x is denoted as: f'(x).
• So we can write:
$$f'(x)~=~\lim_{h\rightarrow 0}{\left[\frac{f(x+h) – f(x)}{h} \right]}$$

---

Now we will see the same solved examples of the previous section:
Solved example 13.6
Find the derivative of f(x) = 3x.
Solution:
• In our present case, f(x) = 3x. So we get:
$\begin{array}{ll}
{}&{f'(x)}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{f(x+h) – f(x)}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{3(x+h) – 3 x}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{3x + 3h – 3x}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{3h}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {3 \lim_{h\rightarrow 0}{\left[\frac{h}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {3 \lim_{h\rightarrow 0}{1}}
&{} \\

{}&{}
& {~=~}& {3 × 1}
&{} \\

{}&{}
& {~=~}& {3}
&{} \\

\end{array}$

Solved example 13.7
Find the derivative of f(x) = 2x² + 3x - 5. Also prove that f'(0) + 3f'(-1) = 0
Solution:
Part (i):
• In our present case, f(x) = 2x² + 3x - 5. So we get:
$\begin{array}{ll}
{}&{f'(x)}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{f(x+h) – f(x)}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{[2(x+h)^2 + 3(x+h) - 5] – [2×(x)^2 + 3 x ~- 5]}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{[2(x^2 + 2hx + h^2) + 3x +3h - 5] – [2 x^2 + 3x - 5]}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{2x^2 + 4hx + 2h^2 + 3x + 3h - 5 – 2 x^2 - 3x + 5}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{4hx + 2h^2 + 3h}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{h(2h +4x + 3)}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{[2h + 4x +3]}}
&{} \\

{}&{}
& {~=~}& {4x + 3}
&{} \\

\end{array}$

Part (ii):
1. We have: f'(x) = 4x + 3
2. Thus we get: f'(0) = [4(0) + 3] = 3
3. Also we get: f'(-1) = [4(-1) + 3] = [-4 + 3] = -1
4. So f'(0) + 3f'(-1) = [3 + 3(-1)] = [3 - 3] = 0

◼ A graphical description can be written in 3 steps:
1. In the fig.13.32 below, f(x) = 2x² + 3x - 5 is plotted in red color.

Fig.13.32

2. We saw that: f'(0) = 3
• That means, the "derivative of f(x)" at (x = 0) is 3.
    ♦ That means, slope of the tangent at (x = 0) is 3.
• When (x=0), f(x) is -5.
    ♦ So we mark the point (0,-5)
• If we have a point and the slope, we can draw a line through that point (recall the slope-point form that we saw in coordinate geometry lessons).
• So we draw a line through (0,-5) at a slope of 3.
    ♦ This line will be the tangent of f(x) at (x=0).
    ♦ This tangent is shown in green color.
3. We saw that: f'(-1) = -1
• That means, the "derivative of f(x)" at (x = -1) is -1.
    ♦ That means, slope of the tangent at (x = -1) is -1.
• When (x=-1), f(x) is -6.
    ♦ So we mark the point (-1,-6)
• If we have a point and the slope, we can draw a line through that point (recall the slope-point form that we saw in coordinate geometry lessons).
• So we draw a line through (-1,-5) at a slope of -1.
    ♦ This line will be the tangent of f(x) at (x=-1).
    ♦ This tangent is shown in white color.

Solved example 13.8
Find the derivative of sin x
Solution:
• In our present case, f(x) = sin x. So we get:
$\begin{array}{ll}
{}&{f'(x)}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{f(x+h) – f(x)}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{[\sin (x+h)] – [sin x]}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{[\sin x \cos h ~+~\cos x \sin h] – [sin x]}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\sin x (\cos h - 1)~+~ \cos x \sin h}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\sin x (\cos h - 1)}{h} \right]}~+~\lim_{h\rightarrow 0}{\left[\frac{\cos x \sin h}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\sin x \lim_{h\rightarrow 0}{\left[\frac{\cos h - 1}{h} \right]}~+~\cos x \lim_{h\rightarrow 0}{\left[\frac{\sin h}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\sin x × 0~+~\cos x × 1}
&{} \\

{}&{}
& {~=~}& {\cos x}
&{} \\

\end{array}$

◼ Let us find the derivatives at some convenient points:
• We have: f'(x) = cos x
• Let us find the derivative at (x = 0) and $\rm{\left(x = \frac{\pi}{3} \right)}$. We get:
    ♦ f'(0) = cos 0 = 1
    ♦ $\rm{f' \left(\frac{\pi}{3}\right)~=~\cos \left(\frac{\pi}{3}\right)~=~\frac{1}{2}}$

◼ A graphical description can be written in 3 steps:
1. In the fig.13.33 below, f(x) = sin x is plotted in red color.

Fig.13.33

2. We saw that: f'(0) = 1
• That means, the "derivative of f(x)" at (x = 0) is 1.
    ♦ That means, slope of the tangent at (x = 0) is 1.
• When (x=0), f(x) is sin 0 = 0.
    ♦ So we mark the point (0,0)
• If we have a point and the slope, we can draw a line through that point (recall the slope-point form that we saw in coordinate geometry lessons).
• So we draw a line through (0,0) at a slope of 1.
    ♦ This line will be the tangent of f(x) at (x=0).
    ♦ This tangent is shown in green color.
3. We saw that: $\rm{f' \left(\frac{\pi}{3}\right)~=~\frac{1}{2}}$
• That means, the "derivative of f(x)" at $\rm{\left(x = \frac{\pi}{3} \right)}$ is $\frac{1}{2}$
    ♦ That means, slope of the tangent at $\rm{\left(x = \frac{\pi}{3} \right)}$ is $\frac{1}{2}$.
• When $\rm{\left(x = \frac{\pi}{3} \right)}$, f(x) is $\frac{\sqrt3}{2}$.
    ♦ So we mark the point $\rm{\left(\frac{\pi}{3}, \frac{\sqrt3}{2} \right)}$
• If we have a point and the slope, we can draw a line through that point (recall the slope-point form that we saw in coordinate geometry lessons).
• So we draw a line through $\rm{\left(\frac{\pi}{3}, \frac{\sqrt3}{2} \right)}$ at a slope of $\frac{1}{2}$.
    ♦ This line will be the tangent of f(x) at $\rm{\left(x = \frac{\pi}{3} \right)}$.
    ♦ This tangent is shown in white color.

Solved example 13.9
Find the derivative of f(x) = 3. Also find f'(0) and f'(3). 
Solution:
Part (i):
• In our present case, f(x) = 3. So we get:

$\begin{array}{ll}
{}&{f'(0)}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{f(x+h) – f(x)}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{[3] – [3]}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{0}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {0}
&{} \\

\end{array}$

Part (ii):
1. We have: f'(x) = 0. This is a constant function.
• So f'(0) = 0.
2. We have: f'(x) = 0. This is a constant function.
• So f'(3) = 0.

◼ In this example, we see that:
    ♦ f’(0) = 0
    ♦ f’(3) = 0
• In fact, we can put any value for a. The derivative will be zero.
• The reason can be written in 2 steps:
1. f(x) = 3 is a constant function.
• That means, whatever be the value of x, the resulting f(x) will be the same.
• If there is no change in f(x), it means that, there is no rate of change.
    ♦ That means, rate of change is zero.
    ♦ That means, derivative is zero.
2. We can think in terms of slope of tangent also.
• For a straight line, the tangent at any point, will be the line itself.
• Here, the straight line is horizontal. So the tangent at any point will also be horizontal.
• For a horizontal line, the slope is zero.

---

• For a given function f, we can find the derivative at every point. The derivative will be a new function denoted by f'.
• In some cases, the derivative will be a constant function.
    ♦ The derivative in solved example 13.6 above is a constant function.
    ♦ The derivative in solved example 13.9 above is a constant function.
• In some cases, the derivative will be an ordinary function.
    ♦ The derivative in solved example 13.7 above is an ordinary function.
    ♦ The derivative in solved example 13.8 above is an ordinary function.

---

◼ Derivative can be denoted in different ways. In the discussions above, we denoted it as f'(x)
• Five more methods are given below:
1. Derivative can be denoted as $\frac{d}{dx}\left(f(x) \right)$.
2. If y = f(x), then the derivative can be denoted as $\frac{dy}{dx}$.
• We can read it in any one of the two ways below:
(i) Derivative of y with respect to x.
(ii) "dy" by "dx"
3. Derivative can be denoted as : $D \left(f(x) \right)$.
4. Derivative of f(x) at (x=a) can be denoted in any one of the three ways below:
(i) $\left.\frac{d}{dx} f(x)\right\vert_{a}$

(ii) $\left.\frac{df}{dx}\right\vert_{a}$

(iii) $\left(\frac{df}{dx}\right)_{x=a}$

---

In the next section, we will see a few more solved examples.

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Chapter 13.12 - Method to Find Derivative

In the previous section, we saw that derivative at a point, is the slope of the tangent at that point. In the section before that, we saw derivative as the rate of change. In this section, we will see an easy method to find the derivative at any given point.

It can be written in 11 steps:
1. In fig.13.30 below, the red curve is the graph of f(x).

Fig.13.30

• Two points P and Q are marked on the red curve.
    ♦ A green horizontal line is drawn through P.
    ♦ A green vertical line is drawn through Q.
    ♦ These horizontal and vertical lines meet at R
• Thus we get a right triangle PQR
2. A green dashed vertical line is drawn through P.
• This vertical line meets the x-axis at (a,0)
• So we can write:
The x-coordinate of P is ‘a’.
• If the x-coordinate is ‘a’, then obviously, the y-coordinate will be f(a)
• Thus we get the coordinates of P: (a,f(a))
3. Another green dashed vertical line is drawn through R.
• This vertical line meets the x-axis at ((a+h),0)
• So we can write:
The x-coordinate of Q is ‘(a+h)’.
• If the x-coordinate is ‘(a+h)’, then obviously, the y-coordinate will be f(a+h)
• Thus we get the coordinates of Q: ((a+h),f(a+h))
4. Using the x-coordinates of P and Q, we can find the horizontal distance between P and Q.
• We get:
Horizontal distance between P and Q = [(a+h) – a] = h
• That means, the length PR = h.
5.  Using the y-coordinates of P and Q, we can find the vertical distance between P and Q.
• We get:
Vertical distance between P and Q = [f(a+h) – f(a)]
• That means, the length QR = [f(a+h) – f(a)].
6. Now we have the base and altitude of the right triangle PQR.
• So we can write the slope of the line PQ.
Slope of PQ = $\frac{[f(a+h) – f(a)]}{h}$
7. If Q is brought very close to P, then the slope calculated in (6) will be the slope of the tangent at P
• For bringing Q closer to P, we must decrease the length h. When h approaches zero, Q will be very close to P.
• h must become very close to zero. At the same time, it must not become exact zero. This can be written as:
$\lim_{h\rightarrow 0} h$
8. From (5), we have the length of the altitude:
[f(a+h) – f(a)]
• When h approaches zero, the base PR will become infinitesimal.
• There will be corresponding changes in the altitude also.
• When h approaches zero, the length of the altitude can be written as: $\lim_{h\rightarrow 0} {[f(a+h) – f(a)]}$
9. So now we can write the slope of the tangent at P:
$\frac{\lim_{h\rightarrow 0} {[f(a+h) – f(a)]}}{\lim_{h\rightarrow 0} h}$
• The limit is being applied to both numerator and denominator. So this can be written in a simplified form as:
Slope of tangent at P = $\lim_{h\rightarrow 0}{\left[\frac{f(a+h) – f(a)}{h} \right]}$
10. But slope of the tangent at P is the derivative at P.
• Point P corresponds to (x = a)
• So we can write:
The result in (9) gives the derivative of f(x) at (x = a).
11. The derivative of f(x) at (x=a) is denoted as: f'(a).
• So we can write:
$$f'(a)~=~\lim_{h\rightarrow 0}{\left[\frac{f(a+h) – f(a)}{h} \right]}$$

---

Now we will see some solved examples:
Solved example 13.6
Find the derivative at x = 2 of the function f(x) = 3x.
Solution:
1. We have: $f'(a)~=~\lim_{h\rightarrow 0}{\left[\frac{f(a+h) – f(a)}{h} \right]}$
2. In our present case, a = 2 and f(x) = 3x.
• So we get:
$\begin{array}{ll}
{}&{f'(2)}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{f(2+h) – f(2)}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{3(2+h) – 3 × 2}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{6 + 3h – 6}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{3h}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {3 \lim_{h\rightarrow 0}{\left[\frac{h}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {3 \lim_{h\rightarrow 0}{1}}
&{} \\

{}&{}
& {~=~}& {3 × 1}
&{} \\

{}&{}
& {~=~}& {3}
&{} \\

\end{array}$

3. We see that the derivative is '3'.
• So we can write:
At the point where (x = 2), the rate of change of y with respect to x is 3.
• In fact, for this problem, the rate of change at every point will be 3.
4. We see that the derivative is '3'.
• So we can write:
At the point where (x = 2), the slope of the tangent is 3.
• In fact, for this problem, the slope of tangent at any point will be 3. The reason can be written in steps:
(i) For a straight line, the tangent at any point, will be the line itself.
(ii) Here, the slope of the line is 3. So the tangent at any point will also have the slope 3

Solved example 13.7
Find the derivative of the function f(x) = 2x² + 3x - 5 at x = -1. Also prove that f'(0) + 3f'(-1) = 0
Solution:
Part (i):
1. We have: $f'(a)~=~\lim_{h\rightarrow 0}{\left[\frac{f(a+h) – f(a)}{h} \right]}$
2. In our present case, a = -1 and f(x) = 2x² + 3x - 5 .
• So we get:
$\begin{array}{ll}
{}&{f'(-1)}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{f(-1+h) – f(-1)}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{[2(-1+h)^2 + 3(-1+h) - 5] – [2×(-1)^2 + 3 × -1 - 5]}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{[2(1 - 2h + h^2) + (-3+3h) - 5] – [2 - 3 - 5]}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{2 - 4h + 2h^2 -3+3h - 5 – 2 + 3 + 5}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{- h + 2h^2}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{h(2h - 1)}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{[2h-1]}}
&{} \\

{}&{}
& {~=~}& {2 × 0~-~1}
&{} \\

{}&{}
& {~=~}& {-1}
&{} \\

\end{array}$

Part (ii):
1. We have: $f'(a)~=~\lim_{h\rightarrow 0}{\left[\frac{f(a+h) – f(a)}{h} \right]}$
2. In our present case, a = 0 and f(x) = 2x² + 3x - 5.
• So we get:
$\begin{array}{ll}
{}&{f'(0)}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{f(0+h) – f(0)}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{[2(0+h)^2 + 3(0+h) - 5] – [2×(0)^2 + 3 × 0 ~- 5]}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{[2h^2 + 3h) - 5] – [2×0 + 3 × 0 ~- 5]}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{2h^2 + 3h - 5 + 5}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{2h^2 + 3h}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{h(2h + 3)}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[2h + 3 \right]}}
&{} \\

{}&{}
& {~=~}& {2 × 0 ~+ 3}
&{} \\

{}&{}
& {~=~}& {3}
&{} \\

\end{array}$

Part (iii):
1. From part (i), we have: f'(-1) = -1
2. From part (ii), we have: f'(0) = 3
3. So we get:
f'(0) + 3f'(-1) = [3 + (3 × -1)] = [3 - 3] = 0

---

At this stage, we are able to understand an important fact. It can be written in 2 steps:
(i) In the previous sections, we saw that:
Limits are subjected to various rules.
(ii) Those rules can be effectively used to evaluate the derivative.

---

Solved example 13.8
Find the derivative of sin x at x = 0
Solution:
1. We have: $f'(a)~=~\lim_{h\rightarrow 0}{\left[\frac{f(a+h) – f(a)}{h} \right]}$
2. In our present case, a = 0 and f(x) = sin x.
• So we get:
$\begin{array}{ll}
{}&{f'(0)}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{f(0+h) – f(0)}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{[\sin (0+h)] – [sin 0]}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{[\sin h] – [sin 0]}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{[\sin h] – [0]}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\sin h}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {1}
&{} \\

\end{array}$

Solved example 13.9
Find the derivative of f(x) = 3 at x = 0 and at x = 3
Solution:
Part (i):
1. We have: $f'(a)~=~\lim_{h\rightarrow 0}{\left[\frac{f(a+h) – f(a)}{h} \right]}$
2. In our present case, a = 0 and f(x) = 3.
• So we get:
$\begin{array}{ll}
{}&{f'(0)}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{f(0+h) – f(0)}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{[3] – [3]}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{0}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {0}
&{} \\

\end{array}$

Part (ii):
1. We have: $f'(a)~=~\lim_{h\rightarrow 0}{\left[\frac{f(a+h) – f(a)}{h} \right]}$
2. In our present case, a = 3 and f(x) = 3.
• So we get:
$\begin{array}{ll}
{}&{f'(3)}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{f(3+h) – f(3)}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{[3] – [3]}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{0}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {0}
&{} \\

\end{array}$

◼ In this example, we see that:
    ♦ f’(0) = 0
    ♦ f’(3) = 0
• In fact, we can put any value for a. The derivative will be zero.
• The reason can be written in 2 steps:
1. f(x) = 3 is a constant function.
• That means, whatever be the value of x, the resulting f(x) will be the same.
• If there is no change in f(x), it means that, there is no rate of change.
    ♦ That means, rate of change is zero.
    ♦ That means, derivative is zero.
2. We can think in terms of slope of tangent also.
• For a straight line, the tangent at any point, will be the line itself.
• Here, the straight line is horizontal. So the tangent at any point will also be horizontal.
• For a horizontal line, the slope is zero.

---

• Now we know how to find the derivative at any given point.
• It would be convenient if we could obtain a general form.
• For example, we saw that:
Derivative of 2x² + 3x -5 at (x = -1) is -1
• If we are asked to find the derivative of this function at say (x = 2), we will have to repeat all the steps using ‘2’.
• If we have a general form, we will be able to quickly find the derivative at any given point.
In the next section, we will see such a general form.

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Chapter 13.11 - Derivative as Slope of Tangent

In the previous section, we saw the basics about derivatives. We saw that, derivative is the rate of change. In this section, we will see another use of derivatives.

It can be written in 9 steps:
1. In fig.13.28 below, the red curve is the graph of a function f(x).

Fig.13.28

• C and D are two adjacent points on the graph.
• A yellow line connects C and D.
    ♦ A vertical green line is drawn through D.
    ♦ A horizontal green line is drawn through C.
• These vertical and horizontal green lines intersect at P.
• So we get a right triangle CPD
• We know that, slope of the line CD will be given by the ratio $\frac{\rm{PD}}{\rm{PC}}$
• The line CD intersects the red curve at two points  C and D
2. Keeping D fixed, let us consider a point which is closer to D
• This new point is marked as C₁ in the fig.13.28
    ♦ A horizontal green line is drawn through C₁
• Thus we get a right triangle C₁P₁D
• So slope of the line C₁D will be given by the ratio $\rm{\frac{P_1 D}{P_1 C_1}}$
• Note that:
    ♦ the denominator of this ratio
    ♦ is smaller than
    ♦ the denominator of the ratio in (1)
3. Keeping D fixed, let us consider a point which is still closer to D
• This new point is marked as C₂ in the fig.13.28
    ♦ A horizontal green line is drawn through C₂
• Thus we get a right triangle C₂P₂D
• So slope of the line C₂D will be given by the ratio $\rm{\frac{P_2 D}{P_2 C_2}}$
• Note that:
    ♦ the denominator of this ratio
    ♦ is smaller than
    ♦ the denominator of the ratio in (2)
4. Keeping D fixed, let us consider a point which is even more closer to D
• This new point is marked as C₃ in the fig.13.28
    ♦ A horizontal green line is drawn through C₃
• Thus we get a right triangle C₃P₃D
• So slope of the line C₃D will be given by the ratio $\rm{\frac{P_3 D}{P_3 C_3}}$
• Note that:
    ♦ the denominator of this ratio
    ♦ is smaller than
    ♦ the denominator of the ratio in (3)
5. Proceeding like this, we can mark points C₄, C₅, C₆, C₇ etc., which are closer and closer to D
• When the point is very close to D, the yellow line will no longer intersect the red curve at two points.
• The yellow line will intersect the red curve only at one point.
• When there is only one point, we can say that:
The yellow line touches the red curve.
• The ‘touch’ occurs at D
6. When the yellow line touches the red curve at D, we say that:
The yellow line is the tangent of the red curve at D
7. Even though there is only a "touch" at one point, we still have a triangle similar to CPD.
Only difference is that, the base of that triangle will be of infinitesimal length.
• Since there is a triangle, we can take the ratio $\rm{\frac{Altitude}{Base}}$. This ratio will give the slope.
• So we can write:
Slope of the tangent = $\frac{\text{PD corresponding to infinitesimal CP}}{\text{infinitesimal CP}}$.
(We use the word “corresponding: because:
For each length of the base CP, there will be a corresponding length for the altitude PD. Even when the base is infinitesimal, there will be a corresponding altitude)
8. But in the previous section, we saw that, the ratio with infinitesimal denominator is called the derivative.
• So we can write:
The slope of the tangent at a point is same as the derivative at that point.
9. We proved that, slope is equal to the derivative. We proved this by taking point C and moving forward towards D
• We can take point E and move backward towards D. We will get the same result. This is shown in fig.13.29 below:

Fig.13.29

• The reader should write all steps related to fig.13.29 and become convinced that slope is same as the derivative.

---

• We saw two applications of derivatives.
(i) Derivative as rate of change.
(ii) Derivative as slope of tangent.
• While discussing rate of change, we did not draw the line CD. Because, for that discussion, such a line is not required.
• But while discussing slope of tangent, we did draw that line. Because, it is that line CD which gradually becomes the tangent at D. This gradual change occurs when the point C becomes closer and closer to D.

---

In the next section, we will see an easy method to find the derivative. 

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Chapter 13.10 - Derivatives

In the previous section, we completed a discussion on limits. In this section, we will see a practical application of limits.

Some basics can be written in 23 steps:

1. Consider an object dropped from the top of a tall cliff.
• The distance (s) traveled by the object in time (t) is given by the formula:
s = 4.9t²
    ♦ distance should be measured in meters
    ♦ time should be measured in seconds.
2. Using this formula, we can get several ordered pairs in the form (t,s):
(i) When t = 0, the distance traveled is zero.
    ♦ Because, s = 4.9 × 0² = 0
    ♦ So we get the ordered pair (t,s) = (0,0)
(ii)  When t = 2, the distance traveled is 19.6 m.
    ♦ Because, s = 4.9 × 2² = 19.6 m
    ♦ So we get the ordered pair (t,s) = A(2,19.6)
(iii)  When t = 4, the distance traveled is 78.4 m.
    ♦ Because, s = 4.9 × 4² = 78.4 m
    ♦ So we get the ordered pair (t,s) = B(4,78.4)
(iv)  When t = 6, the distance traveled is 176.4 m.
    ♦ Because, s = 4.9 × 6² = 176.4 m
    ♦ So we get the ordered pair (t,s) = C(6,176.4)
(v)  When t = 8, the distance traveled is 313.6 m.
    ♦ Because, s = 4.9 × 2² = 313.6 m
    ♦ So we get the ordered pair (t,s) = D(8,313.6)
(vi)  When t = 10, the distance traveled is 490 m.
    ♦ Because, s = 4.9 × 10² = 490 m
    ♦ So we get the ordered pair (t,s) = E(10,490)
(vii)  When t = 12, the distance traveled is 705.6 m.
    ♦ Because, s = 4.9 × 12² = 705.6 m
    ♦ So we get the ordered pair (t,s) = F(12,705.6)
3. We can plot the above ordered pairs on a graph.
    ♦ t should be plotted along the x-axis.
    ♦ s should be plotted along the y-axis.
• Such a graph is shown in fig.13.25 below:

Fig.13.25

• The points representing the ordered pairs are connected using a smooth curve. In the above fig., the smooth curve is drawn in red color.
4. Now begins our task. The task can be explained in 3 steps:
(i) We know that, velocity of a freely falling body would be increasing continuously. It would be a uniformly increasing velocity.
(ii) The task is to find the velocity of the object when t = 8 s
(iii) That is., we want to know the velocity with which the object moves, at the instant when the reading in the stop-watch is 8 s.
5. For this task, we need to concentrate on the region near t = 8 s in the graph.
• That is., we need to concentrate on the region near point D.
• More specifically, we need to concentrate on the region between C and D.
• So we will enlarge the region between C and D. The enlarged portion is shown in fig.13.26 below:

Fig.13.26

6. Let us write two distances:
• From C we know that, when the stop-watch shows 6 s, the distance already fallen is 176.4 m.
• From D we know that, when the stop-watch shows 8 s, the distance already fallen is 313.6 m
7. So the distance fallen in the time duration of 2 s from C to D is (313.6 – 176.4) = 137.2 m
• We know that, average velocity multiplied by the " time duration of travel" will give the distance traveled.
• Let v_av(C-D) be the average velocity with which the object fell from C to D. Then we can write: 137.2 = v_av(C-D) × 2
• From this, we get: $v_{av(C-D)}~=~\frac{137.2}{2}$ = 68.6 m/s
8. Note that, average velocity is a ratio.
• Numerator of the ratio is: Distance fallen.
    ♦ It is the altitude PD of the right triangle CPD in fig.13.26
• Denominator of the ratio is: Time duration in which this distance is fallen.
    ♦ It is the base PC of the right triangle CPD in fig.13.26
◼ So we can write:
$v_{av(C-D)}~=~\frac{PD}{PC}~=~\frac{137.2}{2}$ = 68.6 m/s
9. We have the well known formula to find the velocity at any instant:
v = u + at
• Where:
    ♦ v is the velocity at time t
    ♦ u is the initial velocity
    ♦ a is the acceleration.
• In our present case,
    ♦ u is zero.
    ♦ a is the acceleration due to gravity, which is 9.8 m/s²
• So when time is 8 s, we get: v = 0 + 9.8 × 8 = 78.4 m/s
• This is very different from the average velocity obtained in (7) above.
10. So it seems that, the method of average velocity does not help us to obtain the velocity at D.
• But remember that, the time duration was taken as 2 s. The object would fall a large distance during this time. Consequently the velocity change will also be large.
• Let us try a smaller time duration.
11. Let us consider a new point C₁ between C and D. It corresponds to t = 7 s. This is shown in fig.13.27(a) below:

Fig.13.27

• When t = 7, the distance traveled is 240.1 m.
    ♦ Because, s = 4.9 × 7² = 240.1 m
    ♦ So we get the ordered pair (t,s) = A(7,240.1)
• So the duration from C₁ to D is (8-7) = 1 s
• The distance fallen from C₁ to D is (313.6 – 240.1) = 73.5 m
• Then the average velocity can be obtained as:
$v_{av(C_1 - D)}~=~\frac{P_1 D}{P_1 C_1}~=~\frac{73.5}{1}$ = 73.5 m/s
• This is closer to the actual value of 78.4 m/s that we obtained in (9)
12. Let us decrease the time duration further.
• Consider a new point C₂ between C₁ and D. It corresponds to t = 7.5 s
• When t = 7.5, the distance traveled is 240.1 m.
    ♦ Because, s = 4.9 × 7.5² = 275.6 m
    ♦ So we get the ordered pair (t,s) = A(7.5,275.6)
• So the duration from C₂ to D is (8-7.5) = 0.5 s
• The distance fallen from C₂ to D is (313.6 – 275.6) = 38.0 m
• Then the average velocity can be obtained as:
$v_{av(C_2 - D)}~=~\frac{P_2 D}{P_2 C_2}~=~\frac{38.0}{0.5}$ = 76.0 m/s
• This is even more closer to the actual value of 78.4 m/s that we obtained in (9)
13. Let us decrease the time duration further.
• Consider a new point C₃ between C₂ and D. It corresponds to t = 7.75 s
• When t = 7.75, the distance traveled is 240.1 m.
    ♦ Because, s = 4.9 × 7.5² = 294.3 m
    ♦ So we get the ordered pair (t,s) = A(7.75,294.3)
• So the duration from C₃ to D is (8-7.75) = 0.25 s
• The distance fallen from C₃ to D is (313.6 – 294.3) = 19.3 m
• Then the average velocity can be obtained as:
$v_{av(C_3 - D)}~=~\frac{P_3 D}{P_3 C_3}~=~\frac{19.3}{0.25}$ = 77.2 m/s
• This is even more closer to the actual value of 78.4 m/s that we obtained in (9)
14. Let us write a summary of the above results. It can be written in 5 steps:
(i) The actual velocity at point D is 78.4 m/s
This is obtained from step (9)
(ii) When the time duration from C to D is 2 s, the average velocity is 68.6 m/s
This is obtained from step (7)
(iii) When the time duration from C₁ to D is 1 s, the average velocity is 73.5 m/s
This is obtained from step (11)
(iv) When the time duration from C₂ to D is 0.5 s, the average velocity is 76.0 m/s
This is obtained from step (12)
(v) When the time duration from C₃ to D is 0.25 s, the average velocity is 77.2 m/s
This is obtained from step (13)
15. So it is clear that:
As the time duration decreases, accuracy increases.
16. Note that, time duration is the length of a horizontal line segment.
• In fig.13.27(a), CP, C₁P₁, C₂P₂ and C₃P₃ are the horizontal line segments that we considered. Those line segments have decreasing lengths.
• If we decrease the "horizontal lengths" to a very small value like ‘0.00001’, then we will get a highly accurate result.
• In 0.00001, there are four zeros after the decimal point. There can be a million zeros after the decimal point. Then the length will be very close to zero.
• However, the length must not become exact zero. This is because, division by zero will give a number which is not defined.
• Also, the length cannot become -ve. This is because, the base of a triangle cannot be a -ve value.
• When the length is very close to zero, we say that, it is an infinitesimal length. The dictionary meaning can be seen here.
17. When the base CP (of the triangle CPD) is infinitesimal, we get an accurate result.
• That is:
The ratio $\frac{\text{PD corresponding to infinitesimal CP}}{\text{infinitesimal CP}}$ will give us the exact velocity at D
• The ratio $\frac{\text{PD corresponding to ordinary CP}}{\text{ordinary CP}}$ will give us only the average velocity with which the object travels from C to D.
(We use the word “corresponding: because:
For each length of the base CP, there will be a corresponding length for the altitude PD. Even when the base is infinitesimal, there will be a corresponding altitude)  
18. In the above steps, we considered the portion between C and D.
• We gradually decreased the horizontal distance between C and D. We saw that, by decreasing the horizontal distance to an infinitesimal length, we will get the exact velocity at D
• Just like the portion between C and D, we can consider the portion between D and E also. This is shown in fig.13.27(b) above.
    ♦ First take the points D and E. (time at E is 10 s)
    ♦ Next take D and E₁. (time at E₁ must be 9 s)
    ♦ Next take D and E₂. (time at E₂ must be 8.5 s)
    ♦ Next take D and E₃. (time at E₃ must be 8.25 s)
• In this way, we will be decreasing the horizontal distance between D and E. When the length become infinitesimal, we will get the velocity at D.
• The reader may write all the steps related to fig.13.27(b) and become convinced about this fact.
19. We see that, we can move from both sides.
    ♦ We can move from left towards D.
    ♦ We can move from right towards D.
• Both will give the same value for the velocity at D.
20. The velocity at D is called instantaneous velocity at D.
• It is called "instantaneous" because:
The velocity at D is valid only at the instant when the reading in the stop-watch is 8 s
    ♦ At any instant before 8 s, the velocity will be different.
    ♦ At any instant after 8 s, the velocity will be different.
21. At this stage, we can note an interesting fact. It can be written in steps:
(i) We calculated the velocity as a ratio:
$\frac{\rm{Vertical~distance~between~C~and~D}}{\rm{Horizontal~distance~between~C~and~D}}$
(ii) The vertical distance between C and D is the change in distance. (distance fallen up to D minus distance fallen up to C)
(iii) Horizontal distance between C and D is a time duration.
(iv) That means, we are dividing change in distance by time. So it is the rate of change of displacement.
• Indeed, we know that, velocity is the rate of change of displacement.
• When the denominator becomes infinitesimal, we get the instantaneous velocity.
(v) We can write:
When the denominator is infinitesimal, we are getting the instantaneous rate of change.
22. Remember that, the graph we plotted, is the graph of the function f(t) = s = 4.9t²
• We took a ratio in which:
    ♦ Denominator is infinitesimal.
    ♦ Numerator corresponds to that infinitesimal denominator.
• Such a ratio is called the derivative.
• We can say:
    ♦ The derivative of f(t)
    ♦ at t = 8 s
    ♦ is 78.4 m/s
23. Similarly, t can be any instant shown by the stop-watch.
• At that instant, by taking infinitesimal denominator, we will be able to find the derivative at that instant.
• In our present case, the derivative is the instantaneous velocity.
• In some other case, it will be the amount of water flowing into a reservoir at that instant.
• In some other case, it will be the pressure (at that instant), experienced by a balloon which is being filled with gas.
• Derivatives are very important in many science and engineering topics. In the next few sections, we will see easy methods to find derivatives.

---

In the next section, we will see another application of derivatives. 

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Chapter 13.9 - Solved Examples on Limits

In the previous section, we completed a discussion on sandwich theorem. We also proved an important limit which is: $\lim_{x\rightarrow 0} \frac{\sin x}{x} = 1$
In this section, we will see another important limit.

• We want the limit of $\frac{1 - \cos x}{x}$ when x approaches zero.
• Direct substitution will give $\frac{0}{0}$.
• So we must use an alternate method. It can be written as follows:

$\begin{array}{ll}
{}&{\lim_{x\rightarrow 0} \left[\frac{1 - \cos x}{x} \right]}
& {~=~}& {\lim_{x\rightarrow 0} \left[\frac{1 - \cos x}{x} \times \frac{1 + \cos x}{1 + \cos x}  \right] \color{green}{\text{- - - (I)}}}
&{} \\

{}&{}
& {~=~}& {\lim_{x\rightarrow 0} \left[\frac{1 - \cos^2 x}{x(1 + \cos x)} \right]}
&{} \\

{}&{}
& {~=~}& {\lim_{x\rightarrow 0} \left[\frac{\sin^2 x}{x(1 + \cos x)} \right]}
&{} \\

{}&{}
& {~=~}& {\lim_{x\rightarrow 0} \left[\frac{\sin x}{x} \times \frac{\sin x}{(1 + \cos x)} \right]}
&{} \\

{}&{}
& {~=~}& {\lim_{x\rightarrow 0} \left[\frac{\sin x}{x} \right] \times \lim_{x\rightarrow 0} \left[\frac{\sin x}{(1 + \cos x)} \right]}
&{} \\

{}&{}
& {~=~}& {1 \times \lim_{x\rightarrow 0} \left[\frac{\sin x}{(1 + \cos x)} \right] \color{green}{\text{- - - (II)}}}
&{} \\

{}&{}
& {~=~}& {\lim_{x\rightarrow 0} \left[\frac{\sin x}{(1 + \cos x)} \right]}
&{} \\

{}&{}
& {~=~}& {\frac{\sin 0}{(1 + \cos 0)}}
&{} \\

{}&{}
& {~=~}& {\frac{0}{(1 + 1)}}
&{} \\

{}&{}
& {~=~}& {\frac{0}{2}}
&{} \\

{}&{}
& {~=~}& {0}
&{} \\

\end{array}$

◼ Remarks:
• Line marked as I:
In this line, we multiply both numerator and denominator by (1+cos x)
• Line marked as II:
In this line, we make use of the result: $\lim_{x\rightarrow 0} \frac{\sin x}{x} = 1$

---

• The graph of the function $f(x) = \frac{1 - \cos x}{x}$ is shown in fig.13.23 below:

Fig.13.23

• We see that:
As x approaches zero from either sides, f(x) also approaches zero.

---

So now we have two results:

$(i)~\lim_{x\rightarrow 0} \frac{\sin x}{x} = 1$
$(ii)~\lim_{x\rightarrow 0} \frac{1 - \cos x}{x} = 0$

• These two results can be used to find the limits of complicated trigonometric functions.
• The two results together is known as Theorem 5.

---

Let us see some solved examples:

Solved example 13.4
Evaluate the following:
(i) $\lim_{x\rightarrow 0} \left[\frac{\sin 4x}{\sin 2x} \right]$
(ii) $\lim_{x\rightarrow 0} \left[\frac{\tan x}{x} \right]$
Solution:
Part (i):
$\begin{array}{ll}
{}&{\lim_{x\rightarrow 0} \left[\frac{\sin 4x}{\sin 2x} \right]}
& {~=~}& {\lim_{x\rightarrow 0} \left[\frac{\sin 4x}{\sin 2x} \times \frac{4x}{4x} \times \frac{2x}{2x} \right]}
&{} \\

{}&{}
& {~=~}& {\lim_{x\rightarrow 0} \left[\frac{\sin 4x}{4x} \times \frac{2x}{\sin 2x} \times \frac{4x}{2x} \right]}
&{} \\

{}&{}
& {~=~}& {\lim_{x\rightarrow 0} \left[\frac{\sin 4x}{4x} \times \frac{2x}{\sin 2x} \times 2 \right]}
&{} \\

{}&{}
& {~=~}& {2 \times \lim_{x\rightarrow 0} \left[\frac{\sin 4x}{4x} \times \frac{2x}{\sin 2x}  \right]}
&{} \\

{}&{}
& {~=~}& {2 \times \lim_{x\rightarrow 0} \left[\frac{\sin 4x}{4x} \div \frac{\sin 2x}{2x}  \right]}
&{} \\

{}&{}
& {~=~}& {2 \times \left[ \lim_{x\rightarrow 0} \left[\frac{\sin 4x}{4x} \right] \div  \lim_{x\rightarrow 0} \left[\frac{\sin 2x}{2x}  \right] \right]}
&{} \\

{}&{}
& {~=~}& {2 \times \left[1 \div 1 \right]}
&{} \\

{}&{}
& {~=~}& {2 \times \left[1 \right]}
&{} \\

\end{array}$
 

Alternate method:
$\begin{array}{ll}
{}&{\lim_{x\rightarrow 0} \left[\frac{\sin 4x}{\sin 2x} \right]}
& {~=~}& {\lim_{x\rightarrow 0} \left[\frac{2 \sin 2x \cos 2x}{\sin 2x}  \right]}
&{} \\

{}&{}
& {~=~}& {\lim_{x\rightarrow 0} 2 \cos 2x}
&{} \\

{}&{}
& {~=~}& {2 \cos 2(0)}
&{} \\

{}&{}
& {~=~}& { 2 \cos 0}
&{} \\


{}&{}
& {~=~}& {2 \times 1}
&{} \\

{}&{}
& {~=~}& {2}
&{} \\

\end{array}$

Part (ii):
$\begin{array}{ll}
{}&{\lim_{x\rightarrow 0} \left[\frac{\tan x}{x} \right]}
& {~=~}& {\lim_{x\rightarrow 0} \left[\frac{\sin x}{\cos x \times x} \right]}
&{} \\

{}&{}
& {~=~}& {\lim_{x\rightarrow 0} \left[\frac{\sin x}{x} \times \frac{1}{\cos x} \right]}
&{} \\

{}&{}
& {~=~}& {\lim_{x\rightarrow 0} \left[\frac{\sin x}{x} \right] \times \lim_{x\rightarrow 0} \left[ \frac{1}{\cos x} \right]}
&{} \\

{}&{}
& {~=~}& {1 \times \left[ \frac{1}{\cos 0} \right]}
&{} \\

{}&{}
& {~=~}& {1 \times \left[ \frac{1}{\cos 0} \right]}
&{} \\

{}&{}
& {~=~}& {1 \times \left[ \frac{1}{1} \right]}
&{} \\

{}&{}
& {~=~}& {1}
&{} \\

\end{array}$

Solved example 13.5
Evaluate: $\lim_{x\rightarrow 0} x^2 \cos \left(\frac{1}{x}\right)$
Solution:
1. We know that:
    ♦ cosine of any angle will always be greater than or equal to -1.
    ♦ cosine of any angle will always be less than or equal to 1
• So we can write:
$-1 \le \cos \left(\frac{1}{x} \right) \le 1$
• Multiplying throughout by x², we get:
$-x^2 \le x^2 \cos \left(\frac{1}{x} \right) \le x^2$
2. Consider the result in (1).
• The left term can be written as a function: f(x) = -x²
• The middle term can be written as a function: g(x) = $\rm{x^2 \cos \left(\frac{1}{x} \right)}$
• The right term can be written as a function: h(x) = x²
• So we get: f(x) ≤ g(x) ≤ h(x)
3. We want: $\lim_{x\rightarrow 0} g(x)$
• That is., we want: $\lim_{x\rightarrow 0} \left[\rm{x^2 \cos \left(\frac{1}{x} \right)} \right]$
• Here, direct substitution is not possible because, cosine of 1/0 is not defined.
4. So we try another method. It can be written in 2 steps:
(i) In (2) we saw that, g(x) lies between f(x) and h(x)
• As x approaches zero, f(x) will approach 0
   ♦ That is., $\lim_{x\rightarrow 0} f(x) = 0$  
• As x approaches zero, h(x) will also approach 0
   ♦ That is., $\lim_{x\rightarrow 0} h(x) = 0$  
(ii) We want the limit at zero.
• At that point, the limits of both f(x) and h(x) are 0
• So by sandwich theorem, the limit of g(x) will also be 0
• We can write: $\lim_{x\rightarrow 0} \left[\rm{x^2 \cos \left(\frac{1}{x} \right)} \right] = 0$
5. Fig.13.24 below shows the plot of three functions:
   ♦ f(x) = -x²
   ♦ g(x) = $x^2 \cos \left(\frac{1}{x} \right)$
   ♦ h(x) = x²

Fig.13.24

• We see that, g(x) is sandwiched between f(x) and h(x)
• As x approaches zero, both f(x) and h(x) approach 0.
• So sandwich theorem is indeed applicable.

---

The link below gives a few more solved examples

Exercise 13.1

---

In the next section, we will see derivatives. 

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Chapter 13.8 - The Sandwich Theorem

In the previous section, we saw second theorem. We saw some solved examples also. In this section, we will see the third theorem. Later in this section, we will see the sandwich theorem also.

Theorem 3

This can be written in 5 steps:
1. Consider two functions f and g.
• They are plotted in fig.13.20(a) below:

Fig.13.20

2. Both f and g should have the same domain.
• This can be explained in 2 steps:
(i) We know that, domain of a function is a set.
• This set will contain all input x values that can be used for that function.
(ii) In our present case, both f and g should have the same domain.
• For example,
   ♦ the domain of f consists of all real numbers from -10 to 15,
   ♦ the domain of g must also be the same.
   ♦ -11 cannot be an input x value for g
3. In our present case, functions f and g satisfy a special condition. The condition can be written in 4 steps:
(i) Take any input x value from the domain.
(ii) Calculate f(x) corresponding to that input x.
(iii) Calculate g(x) corresponding to that input x.
(iv) f(x) should be less than g(x).

◼ Mathematically, this condition can be written as:
$f(x) < g(x)$ for all x in the domain
4. The functions in fig.13.20(a), satisfy the above condition in (3).
• We see that:
   ♦ all points of the green curve, lie below the yellow curve.   
5. If the condition in (3) is satisfied, then we get an useful result. It can be written in 4 steps:
(i) Take any input x  (say 'a') from the domain.
(ii) Calculate $\lim_{x\rightarrow a} f(x)$ corresponding to that input x.
• We know that,$\lim_{x\rightarrow a} f(x)$ = f(a)
   ♦ From the fig13.20(a), we see that: f(a) = L₁
(iii) Calculate $\lim_{x\rightarrow a} g(x)$ corresponding to that input x.
• We know that,$\lim_{x\rightarrow a} g(x)$ = g(a)
   ♦ From the fig13.20(a), we see that: g(a) = L₂
(iv) We know that f(a) < g(a). So we get:
$\lim_{x\rightarrow a} f(x)$ will be less than $\lim_{x\rightarrow a} g(x)$
• Indeed, in fig.13.20(a), we  see that, (0,L₁) lies below (0,L₂) 

◼ Mathematically, we can write this as:
$$\lim_{x\rightarrow a} f(x) < \lim_{x\rightarrow a} g(x)$$

---

• The above five steps and fig.13.20(a), completely describes theorem 3.
• Based on theorem 3, we can write theorem 4

Theorem 4
This theorem is also known as sandwich theorem. It can be written in 7 steps:

1. In fig.13.20(b) above, we see that:
f(x) < g(x) < h(x) for every x in the domain.
2. Take any input x  (say 'a') from the domain.
Suppose that:
   ♦ f(a) is equal to L
   ♦ h(a) is also equal to L
3. Given that, g(x) is greater than f(x) for every x
• So g(a) will be greater than f(a)
   ♦ That means, g(a) will be greater than L
4. Given also that, g(x) is lesser than h(x) for every x
• So g(a) will be lesser than h(a)
   ♦ That means, g(a) will be lesser than L
5. Let us compare the above results:
   ♦ In (3), we have: g(a) > L
   ♦ In (4), we have: g(a) < L
• Only one condition can satisfy both the inequalities. That is., g(a) = L
6. Let us write the results in terms of limits:
• If f(a) = L, then we can write: $\lim_{x\rightarrow a} f(x) = L$
• If g(a) = L, then we can write: $\lim_{x\rightarrow a} g(x) = L$
• If h(a) = L, then we can write: $\lim_{x\rightarrow a} h(x) = L$
7. Based on the above steps, we can write:
If $\lim_{x\rightarrow a} f(x) = \lim_{x\rightarrow a} h(x) = l$, then $\lim_{x\rightarrow a} g(x) = L$

---

• The above seven steps and fig.13.20(b), completely describes the sandwich theorem.
• Sandwich theorem can be effectively used to find the limits of complicated functions. All we need are two simple functions such that, the complicated function is sandwiched between the two simple functions. Let us see such an application. It can be written in steps:

1. In fig.13.21(a) below, a unit circle is drawn in red color.

Fig.13.21

• Since it is a unit circle, distances OA = OC = 1 unit
• OC makes an angle of x radians with the X-axis.
• Direction of rotation of OC is anti-clockwise. So angle x is +ve.
• A perpendicular is drawn to the X-axis at A.
   ♦ This perpendicular meets the extension of OC at B.
• A perpendicular is dropped from C on to the X-axis.
   ♦ The foot of this perpendicular is D.
• Line AC is drawn in violet color.  
2. We have to find three areas:
   ♦ Area of triangle OAC
   ♦ Area of sector OAC
   ♦ Area of triangle OAB
3. First we will find the area of triangle OAC. It can be done in 4 steps:
(i) We have:
Area of triangle OAC = $\rm{\frac{1}{2} × OA × CD}$
(ii) Length OA = 1
(iii) In triangle ODC, $\rm{\sin x = \frac{CD}{OC} = \frac{CD}{1} = CD}$
• So CD = sin x
(iv) So from (i), we get:
Area of triangle OAC = $\rm{\frac{1}{2} × 1 × \sin x = \frac{\sin x}{2}}$
4. Next we will find the area of sector OAC. It can be done in 5 steps:
(i) When OC rotates a complete 2π radians, we get the full area of πr² square units.
(ii) So when OC rotates 1 radian, we get an area of:
$\rm{\frac{\pi r^2}{2 \pi} = \frac{r^2}{1} = \frac{1^2}{2} = \frac{1}{2}}$ square units.
(iii) So when OC rotates x radians, we get an area of:
$\rm{x × \frac{1}{2} = \frac{x}{2}}$ square units.
(iv) But area obtained by rotating through x radians is same as the area of the sector OAC.
(v) So we can write:
Area of sector OAC = $\rm{\frac{x}{2}}$ square units.
5. Finally we will find the area of triangle OAB. It can be done in 4 steps:
(i) We have:
Area of triangle OAB = $\rm{\frac{1}{2} × OA × AB}$
(ii) Length OA = 1
(iii) In triangle OAB, $\rm{\tan x = \frac{AB}{OA} = \frac{AB}{1} = AB}$
• So AB = tan x
(iv) So from (i), we get:
Area of triangle OAB = $\rm{\frac{1}{2} × 1 × \tan x = \frac{\tan x}{2}}$
6. Comparing the three areas in the fig.13.21(a), we can see that:
   ♦ Area of triangle OAC is the smallest area.
   ♦ Area of triangle OAB is the largest area.
• So we can write:
Area of △OAC < Area of sector OAC < Area of △OAB
7. Substituting the actual areas, we get:
$\frac{\sin x}{2} < \frac{x}{2}< \frac{\tan x}{2}$
• Multiplying throughout by 2, we get:
$\rm{\sin x < x < \tan x}$
• This can be written as:
$\rm{\sin x < x < \frac{\sin x}{\cos x}}$
• Dividing by sin x, we get:
$\rm{1 < \frac{x}{\sin x} < \frac{1}{\cos x}}$
(For our present discussion, we are considering only those x values from zero to $\frac{\pi}{2}$ radians. For those x values, sin x will be +ve. So dividing by sin x will not alter the nature of the inequalities)
8. Consider the result in (7).
• If we take the reciprocals throughout, the nature of the inequalities will change. We will get:
$\rm{1 > \frac{\sin x}{x} > \frac{\cos x}{1}}$
• This can be rearranged as:
$\rm{\frac{\cos x}{1} < \frac{\sin x}{x} < 1}$
9. Consider the result in (8).
• The right term can be written as a function: f(x) = 1
• The middle term can be written as a function: g(x) = $\rm{\frac{\sin x}{x}}$
• The left term can be written as a function: h(x) = cos x
• So we get: f(x) < g(x) < h(x)
10. We want: $\lim_{x\rightarrow 0} g(x)$
• That is., we want: $\lim_{x\rightarrow 0} \left[\frac{\sin x}{x} \right]$
• Here, direct substitution is not possible because, both numerator and denominator will become zero.
11. So we try another method. It can be written in 2 steps:
(i) In (9) we saw that, g(x) lies between f(x) and h(x)
• As x approaches zero, f(x) will approach 1
   ♦ That is., $\lim_{x\rightarrow 0} f(x) = 1$  
   ♦ Recall that, $\lim_{x\rightarrow 0} 1 =1$.
• As x approaches zero, h(x) will also approach 1
   ♦ That is., $\lim_{x\rightarrow 0} h(x) = 1$  
   ♦ Recall that, $\lim_{x\rightarrow 0} \cos x =1$.
(ii) We want the limit at zero.
• At that point, the limits of both f(x) and h(x) are 1
• So by sandwich theorem, the limit of g(x) will also be 1
• We can write: $\lim_{x\rightarrow 0} \frac{\sin x}{x} = 1$
12. Thus we effectively used the sandwich theorem to find the limit of a complicated function.

---

• Fig.13.22 below shows the plot of three functions:
   ♦ f(x) = cos x
   ♦ g(x) = $\frac{\sin x}{x}$
   ♦ h(x) = 1

Fig.13.22

• Based on the plots, we can write a few more details about the limit of g(x). It can be written in 3 steps:
1. We see that, g(x) is sandwiched between f(x) and h(x)
• As x approaches zero, both f(x) and h(x) approach 1.
• So sandwich theorem is indeed applicable.
2. We used the geometrical construction in fig.13.21(a) to find the limit of $\frac{\sin x}{x}$.
• This fig. helps us to prove the right side limit $\lim_{x\rightarrow 0^{+}} \left[\frac{\sin x}{x} \right]$
• The reason can be explained in 4 steps:
(i) The angle x in fig.13.21(a) is +ve because, it is measured in the anti-clockwise direction.
(ii) So the x values obtained from this fig. are marked on the right side of the origin O.
(iii) We can approach from the right side (starting from say $\frac{\pi}{2}$) towards zero.
(iv) As we approach zero, the angle x decreases. The line OB gets closer and closer to the x-axis.
3. To prove the left side limit, we can use fig.13.21(b)
• The procedure is the same. The reader may write all the necessary steps to find the three areas. The areas can be compared to determine the required limit.
• Some points to remember are:
(i) The angle increases from $\frac{-\pi}{2}$ to zero. The line OB' gets closer and closer to the x-axis.
(ii) The angles are -ve, So sine of those angles will be -ve. (Remember the identity: sin(-x) = -sin x)
(iii) The angles are -ve. So x values are -ve.
(iv) Since the quantities in both (ii) and (iii) are -ve, the ratio $\frac{\sin x}{x}$ will be +ve.
(v) The angles are -ve. So cosine of those angles will be +ve. (Remember the identity: cos(-x) = cos x)

---

• We have completed a discussion on sandwich theorem.
• We also proved an important limit which is: $\lim_{x\rightarrow 0} \frac{\sin x}{x} = 1$
• In the next section, we will see another important limit. 

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Chapter 13.7 - Theorem 2

In the previous section, we saw examples on limits of polynomial functions and rational functions. In this section, we will see the second theorem.

Theorem 2

For any positive integer n,

$$\lim_{x\rightarrow a} \left[\frac{x^n - a^n}{x-a} \right]~=~n a^{n-1}$$

• The expression in this theorem is true even if n is any rational number and a is positive.

• We will see the proof for this theorem in higher classes. At present we will see some solved examples.

Solved example 13.3
(i) Evaluate $\lim_{x\rightarrow a} \left[\frac{x^{15} - 1}{x^{10} - 1} \right]$
(ii) Evaluate $\lim_{x\rightarrow 0} \left[\frac{\sqrt{1+x} - 1}{x} \right]$ 

Solution:
Part (i):
We have:
$\begin{array}{ll}
{}&{\lim_{x\rightarrow 1} \left[\frac{x^{15} - 1}{x^{10} - 1} \right]}
& {~=~}& {\lim_{x\rightarrow 1} \left[\frac{x^{15} - 1}{x - 1}~\div~\frac{x^{10} - 1}{x - 1} \right]}
&{} \\

{}&{}
& {~=~}& {\lim_{x\rightarrow 1} \left[\frac{x^{15} - 1}{x - 1} \right]~\div~\lim_{x\rightarrow 1} \left[\frac{x^{10} - 1}{x - 1} \right]}
&{} \\

{}&{}
& {~=~}& {\left[15 × 1^{15-1}\right]~\div~\left[10 × 1^{10-1}\right]}
&{} \\

{}&{}
& {~=~}& {\left[15 × 1^{14}\right]~\div~\left[10 × 1^{9}\right]}
&{} \\

{}&{}
& {~=~}& {\left[15 × 1\right]~\div~\left[10 × 1\right]}
&{} \\

{}&{}
& {~=~}& {\frac{15}{10}}
&{} \\

{}&{}
& {~=~}& {\frac{3}{2}}
&{} \\

\end{array}$

Part (ii):
1. Let u = x +1
2. We want the limit when x approaches zero.
• Based on (1), we can write:
When x approaches zero, u will approach 1
• Also based on (1), we get: x = u-1
3. So the required limit can be written in another form:
$\lim_{u\rightarrow 1} \left[\frac{\sqrt{u} - 1}{u-1} \right]$
4. Now we can evaluate the limit:
$\begin{array}{ll}
{}&{\lim_{x\rightarrow 0} \left[\frac{\sqrt{1+x} - 1}{x} \right]}
& {~=~}& {\lim_{u\rightarrow 1} \left[\frac{\sqrt{u} - 1}{u-1} \right]}
&{} \\

{}&{}
& {~=~}& {\lim_{u\rightarrow 1} \left[\frac{u^{\frac{1}{2}} - 1}{u-1} \right]}
&{} \\

{}&{}
& {~=~}& {\frac{1}{2} × 1^{\left(\frac{1}{2} - 1\right)}}
&{} \\

{}&{}
& {~=~}& {\frac{1}{2} × 1^{\left(-\frac{1}{2}\right)}}
&{} \\

{}&{}
& {~=~}& {\frac{1}{2} × 1}
&{} \\

{}&{}
& {~=~}& {\frac{1}{2}}
&{} \\

\end{array}$

Alternate method:
$\begin{array}{ll}
{}&{\lim_{x\rightarrow 0} \left[\frac{\sqrt{1+x} - 1}{x} \right]}
& {~=~}& {\lim_{x\rightarrow 0} \left[\frac{\sqrt{1+x} - 1}{x} \times \frac{\sqrt{1+x} + 1}{\sqrt{1+x} + 1}  \right]}
&{} \\

{}&{}
& {~=~}& {\lim_{x\rightarrow 0} \left[\frac{1+x - 1}{x(\sqrt{1+x} + 1)} \right]}
&{} \\

{}&{}
& {~=~}& {\lim_{x\rightarrow 0} \left[\frac{x}{x(\sqrt{1+x} + 1)} \right]}
&{} \\

{}&{}
& {~=~}& {\lim_{x\rightarrow 0} \left[\frac{1}{(\sqrt{1+x} + 1)} \right]}
&{} \\


{}&{}
& {~=~}& {\left[\frac{1}{(\sqrt{1+0} + 1)} \right]}
&{} \\

{}&{}
& {~=~}& {\left[\frac{1}{(1 + 1)} \right]}
&{} \\

{}&{}
& {~=~}& {\frac{1}{2}}
&{} \\

\end{array}$

◼ Remarks:
• In the first line, we multiply both numerator and denominator by "√(1+x) + 1".
    ♦ So the numerator will be in the form (a+b)(a-b)
    ♦ We know that, (a+b)(a-b) = a² - b².

---

In the next section, we will see limits of trigonometric functions. 

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