In the previous section, we saw limits of polynomial functions and rational functions. In this section, we will see some solved examples.
Solved example 13.1
(i) Find the limit of f(x) = x3 - x2 + 1 when x approaches 1
(ii) Find the limit of f(x) = x(x+1) when x approaches 3
(iii) Find the limit of
f(x) = 1 + x + x2 + x3 + . . . + x10 when x approaches -1
Solution:
Part (i):
The given function is a polynomial function. For a polynomial function, the limiting value is same as the value of the function at the required point. So we can write:
$$\lim_{x\rightarrow 1} f(x) = f(1) = (1^3 - 1^2 + 1) = (1 - 1 + 1) = 1$$
Part (ii):
The given function is a polynomial function. For a
polynomial function, the limiting value is same as the value of the
function at the required point. So we can write:
$$\lim_{x\rightarrow 3} f(x) = f(3) = 3(3 + 1) = 3 × 4 = 12$$
Part (iii):
The given function is a polynomial function. For a
polynomial function, the limiting value is same as the value of the
function at the required point. So we can write:
$$\lim_{x\rightarrow -1} f(x) = f(1) = \left[1 + (- 1) + (-1)^2 + (-1)^3 ~+~.~.~.~+~(-1)^{10} \right] = [1 + 1 -1 +1~+~.~.~.~+~1] = 1$$
Note: The sum inside square brackets can be calculated using the formula: $S_n = \frac{a \left(r^n - 1 \right)}{r-1}$
• We saw this in a previous chapter (details here).
• Here, a = 1, r = -1 and n = 11
Solved example 13.2
(i) Find the limit of $f(x) = \frac{x^2 + 1}{x + 100}$ when x approaches 1
(ii) Find the limit of $f(x) = \frac{x^3 - 4x^2 + 4x}{x^2 - 4}$ when x approaches 2
(iii) Find the limit of $f(x) = \frac{x^2 - 4}{x^3 - 4x^2 + 4x}$ when x approaches 2
(iv) Find the limit of $f(x) = \frac{x^3 - 2x^2}{x^2 - 5x + 6}$ when x approaches 2
(v) Find the limit of
$f(x) = \left[\frac{x - 2}{x^2 - x}~-~\frac{1}{x^3 - 3x^2 + 2x}\right]$ when x approaches 1
Solution:
Part (i):
1. The given f(x) is already written in the form $\frac{g(x)}{h(x)}$
2. We see that, both g(x) and f(x) are polynomial functions.
3. So we can write:
$\begin{array}{ll}
{}&{\lim_{x\rightarrow 1} f(x)}
& {~=~}& {\frac{\lim_{x\rightarrow 1} g(x)}{\lim_{x\rightarrow 1} h(x)}}
&{} \\
{}&{}
& {~=~}& {\frac{g(1)}{h(1)}}
&{} \\
{}&{}
& {~=~}& {\frac{1^2 + 1}{1 + 100}}
&{} \\
{}&{}
& {~=~}& {\frac{2}{101}}
&{} \\
\end{array}$
Part (ii):
1. The given f(x) is already written in the form $\frac{g(x)}{h(x)}$
2. We see that, both g(x) and f(x) are polynomial functions.
3. So we can write:
$\begin{array}{ll}
{}&{\lim_{x\rightarrow 2} f(x)}
& {~=~}& {\frac{\lim_{x\rightarrow 2} g(x)}{\lim_{x\rightarrow 2} h(x)}}
&{} \\
{}&{}
& {~=~}& {\frac{g(2)}{h(2)}}
&{} \\
{}&{}
& {~=~}& {\frac{2^3 - 4 × 2^2 + 4 × 2}{2^2 - 4}}
&{} \\
{}&{}
& {~=~}& {\frac{8 - 16 + 8}{4 - 4}}
&{} \\
{}&{}
& {~=~}& {\frac{0}{0}}
&{} \\
\end{array}$
4. Both numerator and denominator are zeroes. So we will factorize the given function. We can write:
$\begin{array}{ll}
{}&{f(x)}
& {~=~}& {\frac{x^3 - 4x^2 + 4x}{x^2 - 4}}
&{} \\
{}&{}
& {~=~}& {\frac{x(x^2 - 4x + 4)}{(x+2)(x-2)}}
&{} \\
{}&{}
& {~=~}& {\frac{x(x-2)^2}{(x+2)(x-2)}}
&{} \\
{}&{}
& {~=~}& {\frac{x(x-2)}{(x+2)}}
&{} \\
\end{array}$
5. Now we can find the limit:
$\begin{array}{ll}
{}&{\lim_{x\rightarrow 2} f(x)}
& {~=~}& {\frac{\lim_{x\rightarrow 2} g(x)}{\lim_{x\rightarrow 2} h(x)}}
&{} \\
{}&{}
& {~=~}& {\frac{g(2)}{h(2)}}
&{} \\
{}&{}
& {~=~}& {\frac{2(2-2)}{(2+2)}}
&{} \\
{}&{}
& {~=~}& {\frac{2 × 0}{4}}
&{} \\
{}&{}
& {~=~}& {\frac{0}{4}}
&{} \\
{}&{}
& {~=~}& {0}
&{} \\
\end{array}$
6. The graph in fig.13.18 below shows that:
When x approaches 2, the value of f(x) indeed approaches zero.
 |
| Fig.13.18 |
7. Note:
This can be written in 3 steps:
(i) In step (4), we canceled (x-2) from both numerator and denominator.
(ii) If we are doing the actual evaluation of f(x) at x = 2, then such a cancellation will not be possible. This is because, if it is an actual evaluation, x will be exactly ‘2’. Then (x-2) will be zero. Division by zero will give a number which does not exist.
(iii) We did the cancellation because, in our present case, it is not an actual evaluation. It is the evaluation of the limiting value. While evaluating the limiting value, x is not exactly equal to 2. So (x-2) will not be zero.
Part (iii):
1. The given f(x) is already written in the form $\frac{g(x)}{h(x)}$
2. We see that, both g(x) and f(x) are polynomial functions.
3. So we can write:
$\begin{array}{ll}
{}&{\lim_{x\rightarrow 2} f(x)}
& {~=~}& {\frac{\lim_{x\rightarrow 2} g(x)}{\lim_{x\rightarrow 2} h(x)}}
&{} \\
{}&{}
& {~=~}& {\frac{g(2)}{h(2)}}
&{} \\
{}&{}
& {~=~}& {\frac{2^2 - 4}{2^3 - 4 × 2^2 + 4 × 2}}
&{} \\
{}&{}
& {~=~}& {\frac{4 - 4}{8 - 16 + 8}}
&{} \\
{}&{}
& {~=~}& {\frac{0}{0}}
&{} \\
\end{array}$
4. Both numerator and denominator are zeroes. So we will factorize the given function. We can write:
$\begin{array}{ll}
{}&{f(x)}
& {~=~}& {\frac{x^2 - 4}{x^3 - 4x^2 + 4x}}
&{} \\
{}&{}
& {~=~}& {\frac{(x+2)(x-2)}{x(x^2 - 4x + 4)}}
&{} \\
{}&{}
& {~=~}& {\frac{(x+2)(x-2)}{x(x-2)^2}}
&{} \\
{}&{}
& {~=~}& {\frac{(x+2)}{x(x-2)}}
&{} \\
\end{array}$
5. Now we can find the limit:
$\begin{array}{ll}
{}&{\lim_{x\rightarrow 2} f(x)}
& {~=~}& {\frac{\lim_{x\rightarrow 2} g(x)}{\lim_{x\rightarrow 2} h(x)}}
&{} \\
{}&{}
& {~=~}& {\frac{g(2)}{h(2)}}
&{} \\
{}&{}
& {~=~}& {\frac{(2+2)}{2(2-2)}}
&{} \\
{}&{}
& {~=~}& {\frac{4}{2 × 0}}
&{} \\
{}&{}
& {~=~}& {\frac{4}{0}}
&{} \\
\end{array}$
6. We can write:
The limit of the given f(x) is not defined when x approaches 2
7. This is evident from the graph in fig.13.19 below:
 |
| Fig.13.19 |
(i) The graph has a particular shape when x is less than zero
(ii) The graph has another shape when x lies between zero and 2
(iii) The graph has yet another shape when x is greater than 2
(iv) We see that:
• When x approaches 2 from the left, f(x) approaches -∞.
• When x approaches 2 from the right, f(x) approaches +∞.
(v) So for this function, the limit when x approaches 2 is not defined.
Part (iv):
1. The given f(x) is already written in the form $\frac{g(x)}{h(x)}$
2. We see that, both g(x) and f(x) are polynomial functions.
3. So we can write:
$\begin{array}{ll}
{}&{\lim_{x\rightarrow 2} f(x)}
& {~=~}& {\frac{\lim_{x\rightarrow 2} g(x)}{\lim_{x\rightarrow 2} h(x)}}
&{} \\
{}&{}
& {~=~}& {\frac{g(2)}{h(2)}}
&{} \\
{}&{}
& {~=~}& {\frac{2^3 - 2 × 2^2}{2^2 - 5 × 2 + 6}}
&{} \\
{}&{}
& {~=~}& {\frac{8 - 8}{4 - 10 + 6}}
&{} \\
{}&{}
& {~=~}& {\frac{0}{0}}
&{} \\
\end{array}$
4. Both numerator and denominator are zeroes. So we will factorize the given function. We can write:
$\begin{array}{ll}
{}&{f(x)}
& {~=~}& {\frac{x^3 - 2x^2}{x^2 - 5x + 6}}
&{} \\
{}&{}
& {~=~}& {\frac{x^2(x - 2)}{(x-3)(x-2)}}
&{} \\
{}&{}
& {~=~}& {\frac{x^2}{(x-3)}}
&{} \\
\end{array}$
5. Now we can find the limit:
$\begin{array}{ll}
{}&{\lim_{x\rightarrow 2} f(x)}
& {~=~}& {\frac{\lim_{x\rightarrow 2} g(x)}{\lim_{x\rightarrow 2} h(x)}}
&{} \\
{}&{}
& {~=~}& {\frac{g(2)}{h(2)}}
&{} \\
{}&{}
& {~=~}& {\frac{2^2}{2-3}}
&{} \\
{}&{}
& {~=~}& {\frac{4}{-1}}
&{} \\
{}&{}
& {~=~}& {-4}
&{} \\
\end{array}$
Part (v):
$f(x) = \left[\frac{x - 2}{x^2 - x}~-~\frac{1}{x^3 - 3x^2 + 2x}\right]$ when x approaches 1
1. First we will write the given function in the form: $f(x) = \frac{g(x)}{h(x)}$
It can be done as follows:
$\begin{array}{ll}
{}&{f(x)}
& {~=~}& {\frac{x - 2}{x^2 - x}~-~\frac{1}{x^3 - 3x^2 + 2x}}
&{} \\
{}&{}
& {~=~}& {\frac{x - 2}{x(x - 1)}~-~\frac{1}{x(x^2 - 3x + 2)}}
&{} \\
{}&{}
& {~=~}& {\frac{(x - 2)(x^2 - 3x + 2)~-~(x-1)}{x(x - 1)(x^2 - 3x + 2)}}
&{} \\
{}&{}
& {~=~}& {\frac{(x - 2)(x - 2)(x-1)~-~(x-1)}{x(x - 1)(x - 2)(x-1)}}
&{} \\
{}&{}
& {~=~}& {\frac{(x-1) \left[(x - 2)(x - 2)~-~1 \right]}{x(x - 1)(x - 2)(x-1)}}
&{} \\
{}&{}
& {~=~}& {\frac{(x-1) \left[(x - 2)^2~-~1 \right]}{x(x - 1)(x - 2)(x-1)}}
&{} \\
{}&{}
& {~=~}& {\frac{(x-1) \left[x^2 - 4x + 4 - 1\right]}{x(x - 1)(x - 2)(x-1)}}
&{} \\
{}&{}
& {~=~}& {\frac{(x-1) \left[x^2 - 4x + 3\right]}{x(x - 1)(x - 2)(x-1)}}
&{} \\
{}&{}
& {~=~}& {\frac{x^2 - 4x + 3}{x(x - 2)(x-1)}}
&{} \\
\end{array}$
2. We see that, both g(x) and f(x) are polynomial functions.
3. So we can write:
$\begin{array}{ll}
{}&{\lim_{x\rightarrow 1} f(x)}
& {~=~}& {\frac{\lim_{x\rightarrow 1} g(x)}{\lim_{x\rightarrow 1} h(x)}}
&{} \\
{}&{}
& {~=~}& {\frac{g(1)}{h(1)}}
&{} \\
{}&{}
& {~=~}& {\frac{1^2 - 4 × 1 + 3}{1(1-2)(1-1)}}
&{} \\
{}&{}
& {~=~}& {\frac{1 - 4 + 3}{1 × -1 × 0}}
&{} \\
{}&{}
& {~=~}& {\frac{0}{0}}
&{} \\
\end{array}$
4. Both numerator and denominator are zeroes. So we will factorize f(x). We can write:
$\begin{array}{ll}
{}&{f(x)}
& {~=~}& {\frac{x^2 - 4x + 3}{x(x - 2)(x-1)}}
&{} \\
{}&{}
& {~=~}& {\frac{(x - 3)(x-1)}{x(x - 2)(x-1)}}
&{} \\
{}&{}
& {~=~}& {\frac{x-3}{x(x-2)}}
&{} \\
\end{array}$
5. Now we can find the limit:
$\begin{array}{ll}
{}&{\lim_{x\rightarrow 1} f(x)}
& {~=~}& {\frac{\lim_{x\rightarrow 1} g(x)}{\lim_{x\rightarrow 1} h(x)}}
&{} \\
{}&{}
& {~=~}& {\frac{g(1)}{h(1)}}
&{} \\
{}&{}
& {~=~}& {\frac{1-3}{1(1-2)}}
&{} \\
{}&{}
& {~=~}& {\frac{-2}{-1}}
&{} \\
{}&{}
& {~=~}& {2}
&{} \\
\end{array}$
6. Note:
This can be written in 3 steps:
(i) In step (4), we canceled (x-1) from both numerator and denominator.
(ii) If we are doing the actual evaluation of f(x) at x = 1,
then such a cancellation will not be possible. This is because, if it
is an actual evaluation, x will be exactly ‘1’. Then (x-1) will be zero.
Division by zero will give a number which does not exist.
(iii) We did the cancellation because, in our present case, it is not an actual evaluation. It is the evaluation of the limiting value. While evaluating the limiting value, x is not exactly equal to 1. So (x-1) will not be zero.
In the next section, we will see the second theorem.
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