Question

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

Answer 1

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

At x=0,the value of the expression is $\frac{0}{0}$ form.

applying L'Hospital Rule in the expression

Differentiate the Numerator and Denominator, and then applying the limit

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

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

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

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

Putting x=0

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

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