Question

$\underset{x\to 0}{lim}\frac{e^x-e^{\sin x}}{x-\sin x}$ equals to

• A. $\frac{1}{2}$
• B. $\frac{1}{4}$
• C. $1$
• D. none of these

Answer 1

Answer: (C)

$1$

Now,

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

Now applying L'Hospital's rule we get,

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

$\left(\frac{0}{0}\right)$ form

Now applying L'Hospital's rule we get,

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

$\left(\frac{0}{0}\right)$ form

Now applying L'Hospital's rule we get,

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

$=\frac{1-1+0+1}{1}$

$=1$.