Question

The value of the limit $\underset{x\to 0}{lim}\frac{e^x-e^{-x}-2x}{x-\sin x}$ is

• A. $4$
• B. $1$
• C. $2$
• D. $\frac{1}{2}$

Answer 1

The correct option is C $2$

${lim}_{x\to 0}\frac{e^x-e^{-x}-2x}{x-sinx}$

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

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

$sinx=x-\frac{x^3}{3!}$

${lim}_{x\to 0}\frac{(1+x+\frac{x^2}{2}+\frac{x^3}{6})-(1-x+\frac{x^2}{2}-\frac{x^3}{6})-2x}{x-(x-\frac{x^3}{6})}$

${lim}_{x\to 0}\frac{2x+\frac{x^3}{3}-2x}{x^{3/6}}$

${lim}_{x\to 0}=2$