Question

If $f\left(x\right)=\frac{x}{\sqrt{x+1}-\sqrt{x}}$ be a real valued function

then which of the following is correct?

• A. $f\left(x\right)$ is continuous, but $f^{{}^{\prime }}\left(0\right)$ does not exist
• B. $f\left(x\right)$ is differentiable at $x=0$
• C. $f\left(x\right)$ is not continuous at $x=0$
• D. $f\left(x\right)$ is not differentiable at $x=0$

Answer 1

Answer: (B)

$f\left(x\right)$ is differentiable at $x=0$

Given $f\left(x\right)=\frac{x}{\sqrt{x+1}-\sqrt{x}}$
Rationalizing it, we get
$f\left(x\right)=x(\sqrt{x+1}+\sqrt{x})$
For $\sqrt{x},x\ge 0$
${lim}_{x\to 0}f\left(x\right)=0$
$f^{\prime }\left(x\right)=(\sqrt{x+1}+\sqrt{x})(1+\frac{1}{2}\sqrt{\frac{x}{x+1}})$
$f^{\prime }\left(0\right)=1$
$f$ is continuous and differentiable at $x=0$