Question

Evaluate: $\underset{x\to 0}{lim}\left[\frac{{(1+x)}^{\frac{1}{8}}-{(1-x)}^{\frac{1}{8}}}{x}\right]$

Answer 1

To evaluate $\underset{x\to 0}{lim}\left[\frac{{(1+x)}^{\frac{1}{8}}-{(1-x)}^{\frac{1}{8}}}{x}\right]$

Put $x=0$

$=\left[\frac{{(1+0)}^{\frac{1}{8}}-{(1-0)}^{\frac{1}{8}}}{0}\right]$

$=\frac{0}{0}$ form.

Now, Use the L-hospital rule

$=\underset{x\to 0}{lim}\left[\frac{{\frac{d}{dx}\left[\right(1+x)}^{\frac{1}{8}}]-{\frac{d}{dx}\left[\right(1-x)}^{\frac{1}{8}}]}{\frac{d}{dx}\left(x\right)}\right]$

$=\underset{x\to 0}{lim}\left[\frac{{\frac{1}{8}\left[\right(1+x)}^{\frac{-7}{8}}(0+1)]-{\frac{1}{8}\left[\right(1-x)}^{\frac{-7}{8}}(0-1)]}{1}\right]$ [$\because \frac{d}{dx}{x}^n=n{x}^{n-1}$]

$=\underset{x\to 0}{lim}\frac{1}{8}(1+x{)}^{\frac{-7}{8}}+\frac{1}{8}(1-x{)}^{\frac{-7}{8}}$

$=\frac{1}{8}(1+0{)}^{\frac{-7}{8}}+\frac{1}{8}(1-0{)}^{\frac{-7}{8}}$

$=\frac{1}{8}+\frac{1}{8}$

$=\frac{2}{8}$

$=\frac{1}{4}$

$\therefore \underset{x\to 0}{lim}\left[\frac{{(1+x)}^{\frac{1}{8}}-{(1-x)}^{\frac{1}{8}}}{x}\right]=\frac{1}{4}$