# How do you find the limit ((1-x)^(1/4)-1)/x as x->0?

Nov 20, 2016

$- \frac{1}{4}$

#### Explanation:

According to the binomial expansion we have

(1-x)^(1/4)=1+1/4(-x)+(1/4(1/4-1))/(2!)(-x)^2+cdots so

((1-x)^(1/4)-1)/x=((1+1/4(-x)+(1/4(1/4-1))/(2!)(-x)^2+cdots)-1)/x
=-1/4+(1/4(1/4-1))/(2!)x+O(x^2)

Here $O \left({x}^{2}\right)$ means an infinite sum of terms with at least order $2$

Then

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

Another approach is rationalizing

$\frac{{\left(1 - x\right)}^{\frac{1}{4}} - 1}{x} = \frac{1}{x} \frac{{\left(1 - x\right)}^{\frac{1}{2}} - 1}{{\left(1 - x\right)}^{\frac{1}{4}} + 1} = \frac{1}{x} \frac{1 - x - 1}{\left({\left(1 - x\right)}^{\frac{1}{4}} + 1\right) \left({\left(1 - x\right)}^{\frac{1}{2}} + 1\right)} = - \frac{1}{\left({\left(1 - x\right)}^{\frac{1}{4}} + 1\right) \left({\left(1 - x\right)}^{\frac{1}{2}} + 1\right)}$ so

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

Nov 20, 2016

Multiply by the conjugate of the numerator over itself.

#### Explanation:

For positive integer $n$, we have

${u}^{n} - {v}^{n} = \left(u - v\right) \left({u}^{n - 1} + {u}^{n - 2} v + {u}^{n - 3} {v}^{2} + \cdot \cdot \cdot + u {v}^{n - 2} + {v}^{n - 1}\right)$.

So

$\left({a}^{\frac{1}{n}} - {b}^{\frac{1}{n}}\right) \left({a}^{\frac{n - 1}{n}} + {a}^{\frac{n - 2}{n}} {b}^{\frac{1}{n}} + {a}^{\frac{n - 3}{n}} {b}^{\frac{2}{n}} + \cdot \cdot \cdot + {a}^{\frac{1}{n}} {b}^{\frac{n - 2}{n}} + {b}^{\frac{n - 1}{n}}\right) = a - b$

${\lim}_{x \rightarrow 0} \frac{{\left(1 - x\right)}^{\frac{1}{4}} - 1}{x} = {\lim}_{x \rightarrow 0} \left(\frac{{\left(1 - x\right)}^{\frac{1}{4}} - 1}{x}\right) \left(\frac{{\left(1 - x\right)}^{\frac{3}{4}} + {\left(1 - x\right)}^{\frac{2}{4}} + {\left(1 - x\right)}^{\frac{1}{4}} + 1}{{\left(1 - x\right)}^{\frac{3}{4}} + {\left(1 - x\right)}^{\frac{2}{4}} + {\left(1 - x\right)}^{\frac{1}{4}} + 1}\right)$

$= {\lim}_{x \rightarrow 0} \frac{\left(1 - x\right) - 1}{x \left({\left(1 - x\right)}^{\frac{3}{4}} + {\left(1 - x\right)}^{\frac{2}{4}} + {\left(1 - x\right)}^{\frac{1}{4}} + 1\right)}$

$= {\lim}_{x \rightarrow 0} \frac{- x}{x \left({\left(1 - x\right)}^{\frac{3}{4}} + {\left(1 - x\right)}^{\frac{2}{4}} + {\left(1 - x\right)}^{\frac{1}{4}} + 1\right)}$

$= {\lim}_{x \rightarrow 0} \frac{- 1}{{\left(1 - x\right)}^{\frac{3}{4}} + {\left(1 - x\right)}^{\frac{2}{4}} + {\left(1 - x\right)}^{\frac{1}{4}} + 1}$

$= \frac{- 1}{{\left(1 - 0\right)}^{\frac{3}{4}} + {\left(1 - 0\right)}^{\frac{2}{4}} + {\left(1 - 0\right)}^{\frac{1}{4}} + 1}$

$= \frac{- 1}{4}$