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## Homework Statement

Proove the statement using the [tex]\epsilon[/tex] and [tex]\delta[/tex] definition of a limit.

The limit of x as it approaches 9 from the left side of the x axis is [tex]\sqrt[4]{9-x}[/tex]=0

## Homework Equations

The limit of x as it approaches a from the left side of the x axis is f(x) = L

if for every number [tex]\epsilon[/tex]>0 there is a number [tex]\delta[/tex]>0 such that if

a-[tex]\delta[/tex]<x<a then |f(x)-L|<[tex]\epsilon[/tex]

## The Attempt at a Solution

9-[tex]\delta[/tex]<x<9 and |[tex]\sqrt[4]{9-x}[/tex]-0|<[tex]\epsilon[/tex]

Using algebra, I've worked my way to |9-x|<[tex]\epsilon^{4}[/tex]. I'm not sure where to go with it now. Do I subtract the 9 over and then change the sign of x, ending up with |x|>-[tex]\epsilon[/tex][tex]^{4}[/tex]+9 ? That doesn't seem right to me.