Question

Prove that $\underset{x\to 0}{lim}\frac{e^x-e^{\sin x}}{x-\sin x}=1$.

Answer 1

$\underset{x\to 0}{lim}\frac{e^x-e^{\sin x}}{x-\sin x}$ $\left(\frac{0}{0}\right)$ form

Now applying L'Hospital's rule we get,

$=\underset{x\to 0}{lim}\frac{e^x-e^{\sin x}\cos x}{1-\cos x}$$\left(\frac{0}{0}\right)$ form

Now applying L'Hospital's rule we get,

$=\underset{x\to 0}{lim}\frac{e^x-e^{\sin x}{\cos }^{2}x+e^{\sin x}\sin x}{\sin x}$$\left(\frac{0}{0}\right)$ form

Now applying L'Hospital's rule we get,

$=\underset{x\to 0}{lim}\frac{e^x-e^{\sin x}{\cos }^{3}x+e^{\sin x}2.\cos x.\sin x+e^{\sin x}{\sin }^{2}x+e^{\sin x}\cos x}{\cos x}=\frac{1-1+0+0+1}{1}=1$