Question

Evaluate the limit:
$\underset{x\to 0}{lim}\frac{\sqrt{1+x}-1}{\log (1+x)}$

Answer 1

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

Rationalising the Numerator
$=\underset{x\to 0}{lim}\frac{\sqrt{1+x}-1}{\log (1+x)}\times \frac{\sqrt{1+x}+1}{\sqrt{1+x}+1}$

$=\underset{x\to 0}{lim}\frac{(\sqrt{1+x}{)}^2-1^2}{\log (1+x)[\sqrt{1+x}+1]}$
$[\because (a+b\left)\right(a-b)=a^2-b^2]$

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

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

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

$=1\times \frac{1}{[\sqrt{1+0}+1]}=\frac{1}{2}[\because \underset{x\to 0}{lim}\frac{\log (1+x)}{x}=1]$

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