True or false; differentiability

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  • #1
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Homework Statement



if g:[-1,1] -> Reals is differentiable with g(0) = 0 and g(x) doesnt equal 0 for x not = 0 and f : Reals -> Reals is a continuous function with f(x)/g(x) ->1 as x->0 then f(x) is differentiable at 0.


Homework Equations





The Attempt at a Solution



I took f(x) = |x| and g(x) = arcsin(x) , does this work as a counter example?
 

Answers and Replies

  • #2
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I took f(x) = |x| and g(x) = arcsin(x) , does this work as a counter example?

No, since the limit

[tex]\lim_{x\rightarrow 1}{\frac{|x|}{arcsin(x)}}[/tex]

doesn't exist.

In fact, let's try to prove this thingy. Can you first work out what f(0) is?
 
  • #3
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its the limit as x --> 0 that i want though?
 
  • #4
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Sorry, typo. That 1 should have been a 0 in the limit...
 
  • #5
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Ok, why doesnt the limit exist? Can i not apply l'Hopitals?
 
  • #7
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hmm thats odd, in my lecture notes i have a definition which says suppose f and g are differentiable at all x in (a,b) \ {c} and g'(x) doesnt = 0 for all x in (a,b)\{c} then
if l = limf'(c)/g'(c) as x--> c exists and = l then it is the limit as x--> x of f(x)/g(x)??

Now im really confused?
 
  • #8
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hmm thats odd, in my lecture notes i have a definition which says suppose f and g are differentiable at all x in (a,b) \ {c} and g'(x) doesnt = 0 for all x in (a,b)\{c} then
if l = limf'(c)/g'(c) as x--> c exists and = l then it is the limit as x--> x of f(x)/g(x)??

Now im really confused?

Yes, I'm sorry, your lecture notes are correct. But the limit still doesn't exist, like you can see in my link...
 
  • #9
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ok. So its true?
 
  • #10
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Try to prove it and find out! :smile:

You'll need to calculate

[tex]\lim_{h\rightarrow 0}{\frac{f(h)-f(0)}{h}}[/tex]

So you'll need to know what f(0) is first...
 
  • #11
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is it 0? since thats the only way we would get f(x) / g(x) -> 1 as x-> 0 ?
 
  • #12
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Yes, it's 0! So you'll only need to calculate

[tex]\lim_{h\rightarrow 0}{\frac{f(h)}{h}}[/tex]

now...
 
  • #13
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ermmm so do i say since f is continuous lim f(h) as h->0 = f(0) = 0
So the whole thing tends to one? argh not sure !?
 
  • #14
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ermmm so do i say since f is continuous lim f(h) as h->0 = f(0) = 0
So the whole thing tends to one? argh not sure !?

f(h) tends to 0, but h also tends to 0, so you have a "0/0" situation. You cannot say that the whole thing tends to 1...
 
  • #15
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L hopitals?
 
  • #16
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No, since you don't know if f is differentiable.

Try to do something with your g...
 
  • #17
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No idea what you mean.. how can i use g to show something about f
 
  • #18
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You could introduce g in that limit somehow.
 

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