Question

$\underset{x\to 0}{lim}\frac{e^{\sin x}-\sin x-1}{x^2}$ is equal to

• A. $\frac{1}{3}$
• B. $1$
• C. $e$
• D. $\frac{1}{2}$

Answer 1

The correct option is D $\frac{1}{2}$

Consider the given function.

$\underset{x\to 0}{lim}\left(\frac{e^{\sin x}-\sin x-1}{x^2}\right)$

This is $\frac{0}{0}$ form.

So, apply L-Hospital rule,

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

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

This is $\frac{0}{0}$ form.

Again, L-Hospital rule

$\underset{x\to 0}{lim}\left(\frac{(e^{\sin x}-1)(-\sin x)+(e^{\sin x}\cos x-0)\cos x}{2}\right)$

$\underset{x\to 0}{lim}\left(\frac{(e^{\sin 0}-1)(-\sin 0)+\left(e^{\sin 0}\cos 0\right)\cos 0}{2}\right)$

$\underset{x\to 0}{lim}\left(\frac{0+(1\times 1)\times 1}{2}\right)$

$=\frac{1}{2}$

Hence, this is the answer.