# How do you prove that the limit of x^(-1/2) = 2 as x approaches 1/4 using the epsilon delta proof?

Nov 10, 2016

Take for example $\delta \left(\epsilon\right) = \min \left(\frac{1}{8} , \frac{3}{32} \epsilon\right)$

#### Explanation:

Choosing $\delta \left(\epsilon\right) = \min \left(\frac{1}{8} , \frac{3}{32} \epsilon\right)$

Take $x : \left\mid x - \frac{1}{4} \right\mid < \delta \setminus \setminus \setminus \implies \frac{1}{8} < x < \frac{3}{8} < 1$

So

$\left\mid \frac{1}{\sqrt{x}} - 2 \right\mid = \left\mid \frac{1 - 2 \sqrt{x}}{\sqrt{x}} \right\mid = \left\mid \frac{1 - 2 \sqrt{x}}{\sqrt{x}} \cdot \frac{1 + 2 \sqrt{x}}{1 + 2 \sqrt{x}} \right\mid =$
$= \left\mid \frac{1 - 4 x}{2 x + \sqrt{x}} \right\mid$

But $\sqrt{x} > x$ when $x < 1$ so

$\left\mid \frac{1}{\sqrt{x}} - 2 \right\mid < \left\mid \frac{1 - 4 x}{3 x} \right\mid < \left\mid \frac{4 \cdot \left(\frac{1}{4} - x\right)}{\frac{3}{8}} \right\mid = \frac{32}{3} \left\mid x - \frac{1}{4} \right\mid < \frac{32}{3} \delta < \frac{32}{3} \cdot \frac{3}{32} \epsilon = \epsilon$