Class example: limit of a function using definition

PirateFan308

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.

1. The problem statement, all variables and given/known data
Using the definition of the limit to show that lim_x→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


2. Relevant 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


3. 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

"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

|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.

PirateFan308

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

Sure, that choice works just as well.

SammyS

Yes, there are many ways to come up with δ .

PirateFan308

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.

Harrisonized

Maybe this will help? It has the definition you're looking for and many examples with solutions. Look at problems 4 and 5.

http://www.math.ucdavis.edu/~kouba/C...ciseLimit.html

SammyS

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

Added in Edit:
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

PirateFan308

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

In the proof you used that d<=1 AND d<=epsilon/5. min(1,epsilon/5) is less than or equal to both of them. d=1 doesn't work if you pick a small epsilon. d=epsilon/5 doesn't work if you pick a large epsilon. Try it.

PirateFan308

Thank you! This makes so much more sense now!