Question

The value of $\underset{x\to 0}{\lim }\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{x}$ is

Answer 1

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