Question

$\underset{x\to 0}{lim}\frac{(1+x{)}^{1/2}-(1-x{)}^{1/2}}{x}$

Answer 1

We have,

${lim}_{x\to 0}\left(\frac{{(1+x)}^{1/2}-{(1-x)}^{1/2}}{x}\right)$

This is the $\frac{0}{0}$ form.

So, apply L-Hospital rule

${lim}_{x\to 0}\left(\frac{\frac{1}{2{(1+x)}^{1/2}}-\frac{1}{2{(1-x)}^{1/2}}(-1)}{1}\right)$

${lim}_{x\to 0}(\frac{1}{2{(1+x)}^{1/2}}+\frac{1}{2{(1-x)}^{1/2}})$

$=\frac{1}{2{(1+0)}^{1/2}}+\frac{1}{2{(1-0)}^{1/2}}$

$=\frac{1}{2}+\frac{1}{2}$

$=1$

Hence, this is the answer.