# True or false; differentiability

## 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.

## The Attempt at a Solution

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

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?

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

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

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

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)

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?

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...

ok. So its true?

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...

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

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

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

now...

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 !?

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...

L hopitals?

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

Try to do something with your g...

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

You could introduce g in that limit somehow.