Proofs on Limit and Derivatives

[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?]

Thanks for your help!

 

[Office_Shredder]

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)

 

[HallsofIvy]

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...]

Let k= 1/h so that you have
\lim_{k\rightarrow \infty} ke^-k[/itex]<br /> and use L&#039;Hopital&#039;s rule.<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> <b>2) Suppose lim (x-a) f(x) = L exists and f(x)<u>&gt;</u>0 for all x not =a. Use the definition of limit to prove that L<u>&gt;</u>0.</b> </div> </div> </blockquote> You mean &quot;limit as x goes to a of f(x)&quot;, not &quot; limit of (x-a)f(x)&quot; surely, since in the second case this is not true. Use &#039;indirect proof&#039;. Suppose L&lt; 0 and take \epsilon= L/2 in the definiton of limit.

 

[kingwinner]

Let k= 1/h so that you have
\lim_{k\rightarrow \infty} ke^-k[/itex]<br /> and use L&#039;Hopital&#039;s rule.<br /> <br /> <br /> You mean &quot;limit as x goes to a of f(x)&quot;, not &quot; limit of (x-a)f(x)&quot; surely, since in the second case this is not true. Use &#039;indirect proof&#039;. Suppose L&lt; 0 and take \epsilon= L/2 in the definiton of limit.

<br /> <br /> 1) But there is an absolute value |h|, letting k=1/h won&#039;t get rid of the absolute value, right?<br /> <br /> <br /> 2) Sorry, that&#039;s a typo...I have corrected it...<br /> 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...

 

[kingwinner]

Can someone please help me? I will be writing my finals tomorrow...