Question

The value of $\underset{x\to 0}{lim}[\frac{a}{x}-\cot \frac{x}{a}]$ is

• A. $0$
• B. $1$
• C. $\text{a}$
• D. $\frac{a}{3}$

Answer 1

The correct option is A $0$

$\underset{x\to 0}{lim}[\frac{a}{x}-\cot \frac{x}{a}]$

$\Rightarrow$ $\underset{x\to 0}{lim}[\frac{a}{x}-\frac{1}{\tan \frac{x}{a}}]$

$\Rightarrow$ $\underset{x\to 0}{lim}\left[\frac{a\tan \frac{x}{a}-x}{x\tan \frac{x}{a}}\right]$

By using L' Hospital Rule,

$\Rightarrow$ $\underset{x\to 0}{lim}\left[\frac{a{\sec }^2\frac{x}{a}.\frac{1}{a}-1}{x{\sec }^2\frac{x}{a}.\frac{1}{a}+\tan \frac{x}{a}}\right]$

$\Rightarrow$ $\underset{x\to 0}{lim}\left[\frac{{\sec }^2\frac{x}{a}-1}{\frac{x}{a}{\sec }^2\frac{x}{a}+\tan \frac{x}{a}}\right]$

$\Rightarrow$ $\underset{x\to 0}{lim}\left[\frac{{\tan }^2\frac{x}{a}}{\frac{x}{a}{\sec }^2\frac{x}{a}+\tan \frac{x}{a}}\right]$

Substituting $x=0$

$\Rightarrow$ $\left[\frac{{\tan }^2\frac{0}{a}}{\frac{0}{a}{\sec }^2\frac{0}{a}+\tan \frac{0}{a}}\right]$

$\Rightarrow$ $\frac{0}{0}$

$\Rightarrow$ $0$