# Class example: limit of a function using definition

I am having trouble understanding how to find the limit of a function (using the definition of a limit). I have a class example, and was wondering if anyone could walk me through the steps.

## Homework Statement

Using the definition of the limit to show that limx→2(x2)=4
f(x) = x2
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| = |x2-4| = |(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:
$\forall$ε>0 $\exists$δ>0 $\forall$x$\in$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?

SammyS
Staff Emeritus
Homework Helper
Gold Member
I am having trouble understanding how to find the limit of a function (using the definition of a limit). I have a class example, and was wondering if anyone could walk me through the steps.

## Homework Statement

Using the definition of the limit to show that limx→2(x2)=4
f(x) = x2
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| = |x2-4| = |(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:
$\forall$ε>0 $\exists$δ>0 $\forall$x$\in$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?
"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?"
Your professor likely did some scratch work, starting with |x2-4|<ε, and then getting his result for δ.​

"in the definition, it says that then |f(x)-ε|<L but we ended up getting |f(x)-L|<ε"

It should be |f(x)-L|<ε in the definition.​

Dick
Homework Helper
|f(x)-epsilon|<L is a typo. |f(x)-L|<epsilon is the correct form. And yes, the professor figured out a delta using the later work and then went back and plugged it in.

Another question, is there more than one δ that will prove this?
Say, Given an arbitrary ε>0, take δ=min{2,ε/6}
If x≠2 and |x-2|<δ, then |x-2|<2 and |x-2|<ε/6
|f(x)-L| = |x2-4| = |(x+2)(x-2)| = |x+2||x-2|
|x-2|<2 => -2<x-2<2 => 0<x<4 => 2<x+2<6 => |x+2|<6
|x-2||x+2| < (6)(ε/6) = ε so |f(x)-L|<ε

Dick
Homework Helper
Another question, is there more than one δ that will prove this?
Say, Given an arbitrary ε>0, take δ=min{2,ε/6}
If x≠2 and |x-2|<δ, then |x-2|<2 and |x-2|<ε/6
|f(x)-L| = |x2-4| = |(x+2)(x-2)| = |x+2||x-2|
|x-2|<2 => -2<x-2<2 => 0<x<4 => 2<x+2<6 => |x+2|<6
|x-2||x+2| < (6)(ε/6) = ε so |f(x)-L|<ε

Sure, that choice works just as well.

SammyS
Staff Emeritus
Homework Helper
Gold Member
Yes, there are many ways to come up with δ .

Why is it that I must say d=min{1,ε/5}? Would it also work if I said that δ=1,ε/5. I'm a bit fuzzy on how the "min" makes this true, or the absence of "min" makes it false.

SammyS
Staff Emeritus
Homework Helper
Gold Member
Why is it that I must say d=min{1,ε/5}? Would it also work if I said that δ=1,ε/5. I'm a bit fuzzy on how the "min" makes this true, or the absence of "min" makes it false.

If ε > 5, then if you say that δ > ε/5, the proof won't work.

Let's say ε = 10.

Then the claim would be that δ = 2 will satisfy the definition.
But if x=3.9, then f(3.99)=15.21, so |f(3.99)-2| = 13.21 > 10

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If ε > 5, then if you say that δ > ε/5, the proof won't work.

So is it standard procedure to always take δ=min if there is more than one condition? Will it ever be wrong for me to make δ=min ?

Dick