Question

Evaluate the limit:

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

Answer 1

At $x\to 0,\frac{\sqrt{1+x}-1}{x}\text{becomes}\frac{0}{0}form$

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

On rationalising numerator, we get

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

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

$[\because (a-b\left)\right(a+b)=a^2-b^2]$

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

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

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

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

$=\frac{1}{2}$

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