Question

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

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

Answer 1

Answer: (B)

$\frac{1}{2}$

We simplify the given expression as,

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

Let $x-sinx=y$

As $x\to 0\text{so does y}\to 0$

Hence, the question transforms into

$(\underset{x\to 0}{lim}\frac{e^{sinx}}{2})\times (\underset{y\to 0}{lim}\frac{e^y-1}{y})$

$=\frac{1}{2}\times 1$

$=\frac{1}{2}$