Class example: limit of a function using definition


by PirateFan308
Tags: class example, definition, limit of a function, limits
PirateFan308
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#1
Nov13-11, 06:12 PM
P: 94
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 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


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?
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SammyS
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Nov13-11, 07:11 PM
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Quote Quote by PirateFan308 View Post
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 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


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?
"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
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#3
Nov13-11, 07:15 PM
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|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
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#4
Nov13-11, 07:19 PM
P: 94

Class example: limit of a function using definition


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
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Nov13-11, 07:27 PM
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Quote Quote by PirateFan308 View Post
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
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Nov13-11, 07:27 PM
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Yes, there are many ways to come up with δ .
PirateFan308
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#7
Nov13-11, 07:33 PM
P: 94
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
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#8
Nov13-11, 07:39 PM
P: 209
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
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Nov13-11, 07:40 PM
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Quote Quote by PirateFan308 View Post
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.

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
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Nov13-11, 07:48 PM
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Quote Quote by SammyS View Post
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
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Nov13-11, 07:50 PM
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Quote Quote by PirateFan308 View Post
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.
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
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#12
Nov14-11, 05:37 PM
P: 94
Thank you! This makes so much more sense now!


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