Question

Evaluate the limit:
$\underset{x\to 0}{lim}\frac{x(e^x-1)}{1-\cos x}$

Answer 1

Given:
$\underset{x\to 0}{lim}\frac{x(e^x-1)}{1-\cos x}$

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

Dividing the numerator and the Denominator $x^2$
$=\underset{x\to 0}{lim}\frac{x(e^x-1)}{x}\times \frac{1}{\frac{2{\sin }^2\frac{x}{2}}{\frac{4x^2}{4}}}$

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

$=\log e\times \frac{2}{1^2}$
$[\because \underset{x\to 0}{lim}\left(\frac{a^x-1}{x}\right)=\log a,\underset{x\to 0}{lim}\frac{\sin x}{x}=1]$

$=1\times 2=2$

Therefore,
$\underset{x\to 0}{lim}\frac{x(e^x-1)}{1-\cos x}=2$