Question

$\underset{x\to 0}{lim}\left[\frac{e^x-e^{\sin x}}{x-\sin x}\right]$ is equal to

Answer 1

Given limit is of $\frac{0}{0}$ at $x=0$, so using L'Hopital's rule:

$\underset{x\to 0}{lim}\frac{e^x-e^{sinx}}{x-sinx}=\underset{x\to 0}{lim}\frac{e^x-e^{sinx}cosx}{1-cosx}=\underset{x\to 0}{lim}\frac{e^x-e^{sinx}co{s}^{2}x+sinx{e}^{sinx}}{sinx}$=

$\underset{x\to 0}{lim}\frac{e^x-e^{sinx}co{s}^{3}x+2cosxsinx{e}^{sinx}+sinxcosx{e}^{sinx}+cosx{e}^{sinx}}{cosx}=\frac{1-1+0+0+1}{1}=1$