Question

${lim}_{x\to 0}\frac{x(e^x-1)}{1-cosx}$

Answer 1

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

$\Rightarrow L={lim}_{x\to 0}\frac{e^x+x{e}^x-1}{+\sin x}$

$={lim}_{x\to 0}\frac{e^x+e^x+x{e}^x}{\cos x}$

$=\frac{e^0+e^0+0.e^0}{\cos 0}$

$=\frac{1+1}{1}$

$=2.$

Hence, the answer is $2.$