Question

Compute $:{lim}_{x\to 0}\frac{e^x-\sin x-1}{x}$

Answer 1

Consider $\underset{x\to 0}{lim}\frac{e^x-\sin x-1}{x}$

Putting $x=0$, we get the limit of $\frac{0}{0}$ form

So, use L' Hosipital rule.

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

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