Proving the Limit of (x^2-1)=3 using Epsilon-Delta Definition

[0range]

Homework Statement

Prove each statement using the epsilon delta definition of limit.

lim (x^2-1)=3
x -> -2

Homework Equations

The Attempt at a Solution

Given E > 0, we need D > 0 such that if |x-(-2)|<D then |(x^2-4|<E.

If |x+2|<1, then -1<x+2<1 -5<x-2<-3 |x-2|<5.

Here's where I'm lost... my answer key says to take D=min{E/5,1} but I don't understand why this is.

Thanks in advance.

 

[LCKurtz]

Homework Statement

Prove each statement using the epsilon delta definition of limit.

lim (x^2-1)=3
x -> -2

Homework Equations

The Attempt at a Solution

Given E > 0, we need D > 0 such that if |x-(-2)|<D then |(x^2-4|<E.

If |x+2|<1, then -1<x+2<1 -5<x-2<-3 |x-2|<5.

Here's where I'm lost... my answer key says to take D=min{E/5,1} but I don't understand why this is.

Thanks in advance.

So if |x+2| < 1 what would be your overestimate for

|x² - 4| = |(x+2)(x-2)| ?

This is what you are trying to make small. How close to -2 does x have to be?

 

[0range]

Sorry... I don't know how to figure that out.

 

[HallsofIvy]

You want to make |(x+2)(x- 2)|&lt; \epsilon. That's the same as |x+ 2|&lt; \epsilon/|x- 2|

Now, you have calculated that -5< x- 2< -3 so that 3<|x- 2< 5. The one you really want is 3< |x- 2|. That way, 1/|x- 2|> 1/3 and so \epsilon/|x-2|&gt; \epsilon/3.

 

[LCKurtz]

So if |x+2| < 1 what would be your overestimate for

|x² - 4| = |(x+2)(x-2)| ?

This is what you are trying to make small. How close to -2 does x have to be?

Sorry... I don't know how to figure that out.

If |x+2| < 1 you have shown that |(x-2)| < 5. So

|x² - 4| = |(x+2)(x-2)| < 5|(x+2)|

How small does |x+2| need to be to make this less than \epsilon? Answer that and you will see where the book's answer comes from.

 

[0range]

I think I get it now... when I'm home from work I'll sit down, work through it and post the answer.

Thanks again, your help is very appreciated guys.