Question

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

Answer 1

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

By using Rationalization

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

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

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

$=\frac{1}{\sqrt{1}+1}=\frac{1}{2}$