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## Homework Statement

Using the definition of the limit to show that lim

_{x→2}(x

^{2})=4

f(x) = x

^{2}

c=2

L=4

Given an arbitrary ε>0, take δ=min{1,ε/5}

If x≠2 and |x-2|<δ then |x-2|<1 and |x-2|< ε/5

|f(x)-L| = |x

^{2}-4| = |(x-2)(x+2)| = |x-2||x+2|

|x-2|<1 => 1<x<3 => 3<x+2<5 => |x+2|<5

|x-2||x+2| < (ε/5)(5) = ε so |f(x)-L|<ε

## Homework Equations

We say that lim f(x)

_{x→c}=L if:

[itex]\forall[/itex]ε>0 [itex]\exists[/itex]δ>0 [itex]\forall[/itex]x[itex]\in[/itex]dom f if x≠c and |x-c|<δ then |f(x)-ε|<L

## The Attempt at a Solution

The biggest thing I am confused about is how the professor got δ? Did he have to do the later work first and then went back and plugged in the answer he got?

Also, in the definition, it says that then |f(x)-ε|<L but we ended up getting |f(x)-L|<ε. Why is this? I understand that we can rearrange the equation, but then doesn't this mess up the absolute value signs?