Question

Evaluate the following limits:  ${\displaystyle \underset{x\to 0}{lim}\left[\frac{\sqrt{1+x^2}-\sqrt{1+x}}{\sqrt{1+x^3}-\sqrt{1+x}}\right]}$

Answer 1

${\displaystyle \underset{x\to 0}{lim}\left[\frac{\sqrt{1+x^2}-\sqrt{1+x}}{\sqrt{1+x^3}-\sqrt{1+x}}\right]}$

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

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

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

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

= ${\displaystyle \underset{x\to 0}{lim}\frac{\sqrt{1+x^3}+\sqrt{1+x}}{(x+1)(\sqrt{1+x^2}+\sqrt{1+x})}}$      ...[∵ x → 0, ∴ x ≠ 0, ∴ x – 1 ≠ 0]

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

= ${\displaystyle \frac{1+1}{1(1+1)}}$

= 1