Question

Evaluate: $\underset{x\to 0}{lim}\frac{e^x-e^{-x}-2x}{x-\sin x}$

• A. 2
• B. 3
• C. 4
• D. 5

Answer 1

The correct option is A 2

Given: $\underset{x\to 0}{lim}\frac{e^x-e^{-x}-2x}{x-\sin x}$

Applying L'Hospital rule $3times$

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

$\Rightarrow {lim}_{x\to 0}\frac{e^x+e^{-x}-2}{1-cosx}$
$\Rightarrow \underset{x\to 0}{lim}\frac{e^x-e^{-x}}{sinx}$

$\Rightarrow \underset{x\to 0}{lim}\frac{e^x+e^{-x}}{cosx}=2$

Hence, option ${}^{\prime }{A}^{\prime }$ is correct.