Question

Evaluate:

$\underset{x\to 0}{lim}\frac{2^x-1}{\sqrt{x+1}-1}$.

Answer 1

Now,

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

Now applying L'Hospital's rule we get,

$=\underset{x\to 0}{lim}\frac{2^x.\log 2}{\frac{1}{2\sqrt{x+1}}}$

$=2\log 2$

$=\log 4$.