Question

Evaluate :
$\underset{x\to 0}{lim}\frac{e^x-e^{\mathrm{sinx}}}{x-\sin x}.$

Answer 1

Solution -

${lim}_{x\to 0}\frac{e^x-e^{sinx}}{x-sinx}$

Apply l-hospital rule -

${lim}_{x\to 0}\frac{e^x+cosxsinx}{1-cosx}$

Apply Again

${lim}_{x\to 0}\frac{e^x-co{s}^{2}x{e}^{sinx+sinx{e}^{sinx}}}{sinx}$

Apply again

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

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

$=1$