Question

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

Answer 1

It is of the Form

$\frac{0}{0}$

So, we can apply L-Hospital Rule

i.e. we will differentiate denominator and numerator separately

$\Rightarrow \frac{\frac{1}{2\sqrt{1+x}}-\frac{-1}{2\sqrt{1-x}}}{2}=\frac{\frac{1}{2\sqrt{1+x}}+\frac{1}{2\sqrt{1-x}}}{2}$

Now, ${lim}_{x\to 0}\frac{\frac{1}{2\sqrt{1+x}}+\frac{1}{2\sqrt{1-x}}}{2}=\frac{\frac{1}{2}+\frac{1}{2}}{2}=\frac{1}{2}$