Question

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

Answer 1

$\underset{x\to 0}{\lim }\left[\frac{\sqrt{1+x}-1}{x}\right]$

It is of the form $\frac{0}{0}$.
Rationalising the numerator:

$\underset{x\to 0}{\lim }\left[\frac{\left(\sqrt{1+x}-1\right)\left(\sqrt{1+x}+1\right)}{x\left(\sqrt{1+x}+1\right)}\right]$

=$\underset{x\to 0}{\lim }\left[\frac{1+x-1}{x\left(\sqrt{1+x}+1\right)}\right]$

= $\frac{1}{\sqrt{1+0}+1}$

= $\frac{1}{2}$