Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Proofs on Limit and Derivatives

  1. May 8, 2007 #1
    1) Prove that f defined by
    f(x)= e^(-1/|x|), x=/=0,
    f(x)= 0, x=0
    is differentiable at 0.


    [I used the definition of derivative
    f'(0)=lim [f(0+h)-f(0)] / h = lim [e^(-1/|h|) / h]
    h->0 h->0
    and I am stuck here and unable to proceed...]


    2) Suppose lim (x->a) f(x) = L exists and f(x)>0 for all x not =a. Use the definition of limit to prove that L>0.

    [when I draw a picture, I can see that this is definitely true, but how can I go about proving it?]

    Thanks for your help!:smile:
     
    Last edited: May 8, 2007
  2. jcsd
  3. May 8, 2007 #2

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    For the second one, start with the definition of limit. Suppose L is less than zero, and show there must be some f(x) for x near a such that f(x) is quite close to L, and hence negative (in more formal terms of course)
     
  4. May 8, 2007 #3

    HallsofIvy

    User Avatar
    Science Advisor

    Let k= 1/h so that you have
    [tex]\lim_{k\rightarrow \infty} ke^-k[/itex]
    and use L'Hopital's rule.

    You mean "limit as x goes to a of f(x)", not " limit of (x-a)f(x)" surely, since in the second case this is not true. Use 'indirect proof'. Suppose L< 0 and take [itex]\epsilon[/itex]= L/2 in the definiton of limit.
     
  5. May 8, 2007 #4
    1) But there is an absolute value |h|, letting k=1/h won't get rid of the absolute value, right?


    2) Sorry, that's a typo...I have corrected it...
    How do you know how epsilon you are going to pick? I am personally having terrible trouble knowing what epsilon to pick to do this kind of proofs...
     
  6. May 8, 2007 #5
    Can someone please help me? I will be writing my finals tomorrow...
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook