Question

Find the limit $\underset{x\to 0}{lim}\frac{x}{\sqrt{1+x}-1}$.

Answer 1

Now,

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

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

$=\underset{x\to 0}{lim}\{\sqrt{1+x}+1\}$

$=2$.