Question

Evaluate the limit:

$\underset{x\to 0}{lim}\frac{x^3\cot x}{1-\cos x}$

Answer 1

We have,

$\underset{x\to 0}{lim}\frac{x^3\cot x}{1-\cos x}$

It becomes 0/0 form.

$\Rightarrow \underset{x\to 0}{lim}\frac{x^3}{\tan x\left(2{\sin }^2\frac{x}{2}\right)}$

$[\because 1-\cos x=2{\sin }^2\frac{x}{2},\cot x=\frac{1}{\tan x}]$

Dividing numerator and denominator by $x^3$, we get

$=\underset{x\to 0}{lim}\frac{1}{\frac{\tan x}{x}\times \frac{2\sin \frac{x}{2}}{\frac{x}{2}}\times \frac{sin\frac{x}{2}}{\frac{x}{2}}\times \frac{1}{4}}$

$=\frac{1}{1\times 2\times 1\times \frac{1}{4}}[\because \underset{x\to 0}{lim}\frac{\sin x}{x}=1,\underset{x\to 0}{lim}\frac{\tan x}{x}=1]$

$=2$

Therefore,

$\underset{x\to 0}{lim}\frac{x^3\cot x}{1-\cos x}=2$