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
In the next section, we will see derivatives.
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