True or false; differentiability

[stukbv]

Homework Statement

if g:[-1,1] -> Reals is differentiable with g(0) = 0 and g(x) doesn't 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?

 

[micromass]

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

No, since the limit

$$\lim_{x\rightarrow 1}{\frac{|x|}{arcsin(x)}}$$

doesn't exist.

In fact, let's try to prove this thingy. Can you first work out what f(0) is?

 

[stukbv]

its the limit as x --> 0 that i want though?

 

[micromass]

Sorry, typo. That 1 should have been a 0 in the limit...

 

[stukbv]

Ok, why doesn't the limit exist? Can i not apply l'Hopitals?

 

[micromass]

You can only use l'Hopital when both functions are differentiable at 0. And |x| is not.
If you graph the function (see http://www.wolframalpha.com/input/?i=abs(x)/arcsin(x) ) then you'll see that the limit doesn't exist (i.e. the left and right limits do not agree)

 

[stukbv]

hmm that's 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) doesn't = 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 I am really confused?

 

[micromass]

hmm that's 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) doesn't = 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 I am 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...

 

[stukbv]

ok. So its true?

 

[micromass]

Try to prove it and find out!

You'll need to calculate

$$\lim_{h\rightarrow 0}{\frac{f(h)-f(0)}{h}}$$

So you'll need to know what f(0) is first...

 

[stukbv]

is it 0? since that's the only way we would get f(x) / g(x) -> 1 as x-> 0 ?

 

[micromass]

Yes, it's 0! So you'll only need to calculate

$$\lim_{h\rightarrow 0}{\frac{f(h)}{h}}$$

now...

 

[stukbv]

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 !?

 

[micromass]

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...

 

[stukbv]

L hopitals?

 

[micromass]

No, since you don't know if f is differentiable.

Try to do something with your g...

 

[stukbv]

No idea what you mean.. how can i use g to show something about f

 

[micromass]

You could introduce g in that limit somehow.