1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Class example: limit of a function using definition

  1. Nov 13, 2011 #1
    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?
     
  2. jcsd
  3. Nov 13, 2011 #2

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    "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.​
     
  4. Nov 13, 2011 #3

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    |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.
     
  5. Nov 13, 2011 #4
    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
     
  6. Nov 13, 2011 #5

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Sure, that choice works just as well.
     
  7. Nov 13, 2011 #6

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Yes, there are many ways to come up with δ .
     
  8. Nov 13, 2011 #7
    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.
     
  9. Nov 13, 2011 #8
    Last edited: Nov 13, 2011
  10. Nov 13, 2011 #9

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    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
     
    Last edited: Nov 13, 2011
  11. Nov 13, 2011 #10
    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 ?
     
  12. Nov 13, 2011 #11

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
  13. Nov 14, 2011 #12
    Thank you! This makes so much more sense now!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?