# Proofs on Limit and Derivatives

1. May 8, 2007

### kingwinner

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?]

Last edited: May 8, 2007
2. May 8, 2007

### Office_Shredder

Staff Emeritus
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)

3. May 8, 2007

### HallsofIvy

Staff Emeritus
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 $\epsilon$= L/2 in the definiton of limit.

4. May 8, 2007

### kingwinner

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...

5. May 8, 2007