Question

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

• A. $1$
• B. $-1$
• C. $2$
• D. $0$

Answer 1

The correct option is C $2$

Let, $L=\underset{x\to 0}{lim}\frac{e^x-e^{-x}-2x}{x-\sin x}$
Using series expansion,

$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+................$

$e^{-x}=1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+................$

and, $\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+..............$

$\therefore L=\underset{x\to 0}{lim}\frac{2(\frac{x^3}{3!}+\frac{x^5}{5!}+...........)}{\frac{x^3}{3!}-\frac{x^5}{5!}+\frac{x^7}{7!}+......}$

$=\underset{x\to 0}{lim}\frac{2(\frac{1}{3!}+\frac{x^2}{5!}+...........)}{\frac{1}{3!}-\frac{x^2}{5!}+\frac{x^7}{7!}+......}=2$