Question

Evaluate the limit:
$\underset{x\to 0}{lim}\frac{\sqrt{1+x^2}-\sqrt{1+x}}{\sqrt{1+x^3}-\sqrt{1+x}}$

Answer 1

At $x\to 0$,

$\frac{\sqrt{1+x^2}-\sqrt{1+x}}{\sqrt{1+x^3}-\sqrt{1+x}}$ becomes $\frac{0}{0}$

$\underset{x\to 0}{lim}\frac{\sqrt{1+x^2}-\sqrt{1+x}}{\sqrt{1+x^3}-\sqrt{1+x}}$

On rationalising numerator and denominator, we get

$=\underset{x\to 0}{lim}\frac{(\sqrt{1+x^2}-\sqrt{1+x})(\sqrt{1+x^2}+\sqrt{1+x})(\sqrt{1+x^3}+\sqrt{1+x})}{(\sqrt{1+x^3}-\sqrt{1+x})(\sqrt{1+x^2}+\sqrt{1+x})(\sqrt{1+x^3}+\sqrt{1+x})}$

$=\underset{x\to 0}{lim}\frac{\left(\right(\sqrt{1+x^2}{)}^2-\left(\sqrt{1+x}{)}^2\right)(\sqrt{1+x^3}+\sqrt{1+x})}{\left(\right(\sqrt{1+x^3}{)}^2-\left(\sqrt{1+x}{)}^2\right)(\sqrt{1+x^2}+\sqrt{1+x})}$

$=\underset{x\to 0}{lim}\frac{(1+x^2-1-x)(\sqrt{1+x^3}+\sqrt{1+x})}{(1+x^3-1-x)(\sqrt{1+x^2}+\sqrt{1+x})}$

$=\underset{x\to 0}{lim}\frac{(x^2-x)(\sqrt{1+x^3}+\sqrt{1+x})}{(x^3-x)(\sqrt{1+x^2}+\sqrt{1+x})}$

$=\underset{x\to 0}{lim}\frac{x(x-1)(\sqrt{1+x^3}+\sqrt{1+x})}{x(x^2-1)(\sqrt{1+x^2}+\sqrt{1+x})}$

$=\underset{x\to 0}{lim}\frac{x(x-1)(\sqrt{1+x^3}+\sqrt{1+x})}{x(x-1)(1+x)+(\sqrt{1+x^2}+\sqrt{1+x})}$

$[x\ne 0,(x-1)\ne 0]$

$=\underset{x\to 0}{lim}\frac{\sqrt{1+x^3}+\sqrt{1+x}}{(x+1)(\sqrt{1+x^2}+\sqrt{1+x})}$

$=\frac{(\sqrt{1+0^3}+\sqrt{1+x})}{(0+1)(\sqrt{1+0^2}+\sqrt{1+x})}$

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