# Class example: limit of a function using definition

 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 3 |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: $\forall$ε>0 $\exists$δ>0 $\forall$x$\in$dom f if x≠c and |x-c|<δ then |f(x)-ε|
Emeritus
HW Helper
PF Gold
P: 7,808
 Quote by 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 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 3 |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: $\forall$ε>0 $\exists$δ>0 $\forall$x$\in$dom f if x≠c and |x-c|<δ then |f(x)-ε|
"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.
 Sci Advisor HW Helper Thanks P: 25,228 |f(x)-epsilon|
 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 0 2 |x+2|<6 |x-2||x+2| < (6)(ε/6) = ε so |f(x)-L|<ε
HW Helper
Thanks
P: 25,228
 Quote by 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 0 2 |x+2|<6 |x-2||x+2| < (6)(ε/6) = ε so |f(x)-L|<ε
Sure, that choice works just as well.
 Emeritus Sci Advisor HW Helper PF Gold P: 7,808 Yes, there are many ways to come up with δ .
 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.
 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
Emeritus
HW Helper
PF Gold
P: 7,808
 Quote by 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.
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
P: 94
 Quote by SammyS 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 ?