Question

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

• A. $0$
• B. $1$
• C. $-1$
• D. none of these

Answer 1

The correct option is B ($1$)

Given $\underset{x\to 0}{lim}\frac{e^{\sin x}-1}{x}$
Since, we have a formula $\underset{x\to 0}{lim}\frac{e^x-1}{x}=1$
Since as $x\to 0$ implies $\sin x\to 0$ as well.
$\Rightarrow \underset{\sin x\to 0}{lim}\frac{e^{\sin x}-1}{\sin x}=1$
To get into above form, multiply given expression by$\frac{\sin x}{\sin x}$
$=\underset{x\to 0}{lim}\frac{e^{\sin x}-1}{x}\times \frac{\sin x}{\sin x}$

$=\underset{x\to 0}{lim}\frac{\frac{e^{\sin x}-1}{\sin x}}{\frac{x}{\sin x}}$
$=\underset{x\to 0}{lim}\frac{e^{\sin x}-1}{\sin x}\times \underset{x\to 0}{lim}\frac{1}{\frac{x}{\sin x}}$
$=\underset{\sin x\to 0}{lim}\frac{e^{\sin x}-1}{\sin x}\times \underset{x\to 0}{lim}\frac{\sin x}{x}$
$=1\times 1$ [Since, from above formula and $\underset{x\to 0}{lim}\frac{\sin x}{x}=1$]

$=1$.