Question

The function $f\left(x\right)=\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{x^2}$ for $x\ne 0;$ $f\left(0\right)=1$ at $x=0$ is

• A. continuous
• B. discontinuous
• C. not determined
• D. none

Answer 1

Answer: (B)

discontinuous

$f\left(x\right)=\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{x^2}lefthandlimit=\underset{x\to 0^{-}}{limit}f\left(x\right)\underset{x\to 0^{-}}{limit}\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{x^2}\times \frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}=\underset{x\to 0^{-}}{limit}\frac{1+x^2-(1-x^2)}{x^2}=\underset{x\to 0^{-}}{limit}\frac{2x^2}{x^2}=2Righthandlimit=\underset{x\to 0^{+}}{limit}f\left(x\right)\underset{x\to 0^{+}}{limit}\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{x^2}\times \frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}=\underset{x\to 0^{+}}{limit}\frac{1+x^2-(1-x^2)}{x^2}=\underset{x\to 0^{+}}{limit}\frac{2x^2}{x^2}=2\underset{x\to 0^{-}}{limit}f\left(x\right)=\underset{x\to 0^{+}}{limit}f\left(x\right)\ne f\left(0\right)$

So this function is discontinuous at $x=0$

Correct option is B