Question

Using L' Hospital's Rule, evaluate
$\underset{x\to 0}{lim}\left(\frac{e^x-e^{-x}-2x}{x-\sin x}\right)$

Answer 1

$\underset{x\to 0}{lim}\left(\frac{e^x-e^{-x}-2x}{x-\sin x}\right)\left[As\left(\frac{0}{0}\right)\right]$ form

Using L'Hospital's Rule,

$\underset{x\to 0}{lim}\frac{e^x+e^{-x}-2}{1-\cos x}\left[As\left(\frac{0}{0}\right)\right]$ form

Again using L' Hospital's Rule.

$\underset{x\to 0}{lim}\frac{e^x-e^{-x}}{\sin x}$, $As\left[\left(\frac{0}{0}\right)\right]$ form

Once again using L' Hospital's Rule,

$\Rightarrow \underset{x\to 0}{lim}\frac{e^x+e^{-x}}{\cos x}=\frac{1+1}{1}=2$