Register to reply

Class example: limit of a function using definition

Share this thread:
PirateFan308
#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?
Phys.Org News Partner Science news on Phys.org
Experts defend operational earthquake forecasting, counter critiques
EU urged to convert TV frequencies to mobile broadband
Sierra Nevada freshwater runoff could drop 26 percent by 2100
SammyS
#2
Nov13-11, 07:11 PM
Emeritus
Sci Advisor
HW Helper
PF Gold
P: 7,819
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
#3
Nov13-11, 07:15 PM
Sci Advisor
HW Helper
Thanks
P: 25,228
|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
#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
#5
Nov13-11, 07:27 PM
Sci Advisor
HW Helper
Thanks
P: 25,228
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
#6
Nov13-11, 07:27 PM
Emeritus
Sci Advisor
HW Helper
PF Gold
P: 7,819
Yes, there are many ways to come up with δ .
PirateFan308
#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
#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
#9
Nov13-11, 07:40 PM
Emeritus
Sci Advisor
HW Helper
PF Gold
P: 7,819
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
#10
Nov13-11, 07:48 PM
P: 94
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
#11
Nov13-11, 07:50 PM
Sci Advisor
HW Helper
Thanks
P: 25,228
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
#12
Nov14-11, 05:37 PM
P: 94
Thank you! This makes so much more sense now!


Register to reply

Related Discussions
Proving a function is not continuous using epsilon and delta definition of a limit Calculus & Beyond Homework 3
How to prove definition of exponential function as limit of powers converges Calculus 3
Limit Of A Function From Definition Calculus 7
Help with epsilon-delta definition of a limit of a function Calculus & Beyond Homework 3
Help with formal definition of the limit of a function General Math 9