Class example: limit of a function using definitionby PirateFan308 Tags: class example, definition, limit of a function, limits 

#1
Nov1311, 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 lim_{x→2}(x^{2})=4 f(x) = x^{2} c=2 L=4 Given an arbitrary ε>0, take δ=min{1,ε/5} If x≠2 and x2<δ then x2<1 and x2< ε/5 f(x)L = x^{2}4 = (x2)(x+2) = x2x+2 x2<1 => 1<x<3 => 3<x+2<5 => x+2<5 x2x+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 xc<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? 



#2
Nov1311, 07:11 PM

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Your professor likely did some scratch work, starting with x^{2}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"




#3
Nov1311, 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.




#4
Nov1311, 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 x2<δ, then x2<2 and x2<ε/6 f(x)L = x^{2}4 = (x+2)(x2) = x+2x2 x2<2 => 2<x2<2 => 0<x<4 => 2<x+2<6 => x+2<6 x2x+2 < (6)(ε/6) = ε so f(x)L<ε 



#5
Nov1311, 07:27 PM

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#6
Nov1311, 07:27 PM

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




#7
Nov1311, 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.




#8
Nov1311, 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 



#9
Nov1311, 07:40 PM

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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 



#10
Nov1311, 07:48 PM

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#11
Nov1311, 07:50 PM

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#12
Nov1411, 05:37 PM

P: 94

Thank you! This makes so much more sense now!



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