Question

Solve $\underset{x\to 0}{lim}\frac{\cos ecx-\cot x}{x}$ is

Answer 1

We have,

$\underset{x\to 0}{lim}\left(\frac{\cos ecx-\cot x}{x}\right)$

This is $\frac{0}{0}$ form.

So, apply L-Hospital rule,

$\underset{x\to 0}{lim}\left(\frac{-\cos ecx\cot x-(-\cos e{c}^{2}x)}{1}\right)$

$\underset{x\to 0}{lim}(\cos e{c}^{2}x-\cos ecx\cot x)$

$\underset{x\to 0}{lim}(\frac{1}{{\sin }^{2}x}-\frac{\cos x}{{\sin }^{2}x})$

$\underset{x\to 0}{lim}\left(\frac{1-\cos x}{{\sin }^{2}x}\right)$

This is also $\frac{0}{0}$ form.

So, again apply L-Hospital rule,

$\underset{x\to 0}{lim}\left(\frac{0-(-\sin x)}{2\sin x\cos x}\right)$

$\underset{x\to 0}{lim}\left(\frac{\sin x}{2\sin x\cos x}\right)$

$\underset{x\to 0}{lim}\left(\frac{1}{2\cos x}\right)$

$=\frac{1}{2\cos 0}$

$=\frac{1}{2}$

Hence, the value is $\frac{1}{2}$.