# How do you find lim ((1-x)^(1/4)-1)/x as x->0 using l'Hospital's Rule?

Apr 14, 2017

$- \frac{1}{4}$

#### Explanation:

Take the derivative of the top and bottom. Don't forget chain rule:

${\lim}_{x \rightarrow 0} \left(\frac{1}{4}\right) \left(- 1\right) {\left(1 - x\right)}^{- \frac{3}{4}} / 1$

Plug in $0$ for $x$:

$\left(\frac{1}{4}\right) \left(- 1\right) {\left(1\right)}^{- \frac{3}{4}} / 1 = - \frac{1}{4}$

Apr 14, 2017

Alt approach:

${\lim}_{x \to 0} \frac{{\left(1 - x\right)}^{\frac{1}{4}} - 1}{x}$

$= {\lim}_{x \to 0} \frac{\left(1 - \frac{1}{4} x + O \left({x}^{2}\right)\right) - 1}{x}$

$= {\lim}_{x \to 0} - \frac{1}{4} + O \left(x\right) = - \frac{1}{4}$