Question

Evaluate the following limits:

${lim}_{x\to 0}\frac{\sqrt{1+x+x^2}-1}{x}$

Answer 1

${lim}_{x\to 0}\frac{\sqrt{1+x+x^2}-1}{x}$

Rationalising the numerator

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

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

$={lim}_{x\to 0}\left[\frac{x(1+x)}{x(\sqrt{1+x+x^2}+1)}\right]={lim}_{x\to 0}\left[\frac{x(1+x)}{\sqrt{1+x+x^2}+1}\right]$

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

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