Question

Find the value of
$\underset{x\to 0}{lim}\frac{x(e^x-1)}{1-\cos x}$

Answer 1

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

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

$=\underset{x\to 0}{lim}\left(\frac{(e^x-1)}{x}\right).2.\frac{x^2/4}{{\sin }^{2}x/2}=\underset{x\to 0}{lim}2\left(\frac{(e^x-1)}{x}\right).2.{\left(\frac{x^2/2}{{\sin }^{2}x/2}\right)}^2$

$=2\left(1\right)\left(1\right)=2$