- #1

- 65

- 2

## Homework Statement

Prove [tex]lim_{x->4}\frac{x-4}{\sqrt{x}-2}=4[/tex]

## Homework Equations

Epsilon\delta definition

## The Attempt at a Solution

I can see that a direct evaluation at 4 leads to an indeterminate form, so:

[tex]\frac{x-4}{\sqrt{x}-2}*\frac{\sqrt{x}+2}{\sqrt{x}+2} \; \mbox{simplifies to} \; \sqrt{x}+2 \; \mbox{when} \; x \ne 4[/tex]

Via epsilon\delta definition,

[tex]|\sqrt{x}+2-4|<\epsilon \; \mbox{whenever} \; |x-4|<\delta[/tex]

Expanding the right side,

[tex]-\delta<x-4<\delta[/tex]

[tex]-\delta+4<x<\delta+4[/tex]

[tex]\sqrt{-\delta+4}<\sqrt{x}<\sqrt{\delta+4} \; \mbox{when} \; \delta \leq 4[/tex]

[tex]\sqrt{-\delta+4}-2<\sqrt{x}-2<\sqrt{\delta+4}-2[/tex]

Thus, [tex]\delta=min\{\sqrt{-\delta+4}-2,4,\sqrt{\delta+4}-2\}[/tex]

Given this definition of delta, it is elementary to work backwards towards [tex]|\sqrt{x}+2|<\delta[/tex]

Have I made any errors in my proof?