Question

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

Answer 1

${lim}_{x\to 0}\frac{x(e^x-1)}{1-\cos x}={lim}_{x\to }\frac{x(e^x-1)}{2{\sin }^2\left(\frac{x}{2}\right)}={lim}_{x\to 0}\frac{(e^x-1)}{2x}{lim}_{x\to 0}\frac{4}{\left(\frac{\sin \left(\frac{x}{2}\right)}{\frac{x}{2}}\right)}=\frac{1}{2}\times 4=2$