Question

Evaluate:

$\underset{x\to 0}{lim}\frac{e^x-e^{-x}-2\log (1+x)}{x\sin x}$.

Answer 1

Now,

$\underset{x\to 0}{lim}\frac{e^x-e^{-x}-2\log (1+x)}{x\sin x}$ $\left(\frac{0}{0}\right)$ form.

Now applying L'Hospital's rule we get,

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

$\left(\frac{0}{0}\right)$ form.

Now applying L'Hospital's rule we get,

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

$=\frac{2}{2}$

$=1$.