# Limit of differentiable functions question

1. Mar 13, 2009

### transgalactic

f(x) and g(x) are differentiable on 0
f(0)=g(0)=0
$$\lim _{x->0}\frac{cos(f(x))-cos(g(x))}{x^2}=\lim _{x->0}\frac{-2sin(\frac{f(x)+g(x)}{2})sin(\frac{f(x)-g(x)}{2})}{x^2}=-2$$

because i can use (sin x)/x=1 here
is it ok??

2. Mar 13, 2009

### HallsofIvy

No, you don't have "sin(x)/x", you have
$$\frac{sin(\frac{f(x)+ g(x)}{2})}{x^2}[/itex] whether this converges at all depends on how fast f(x) and g(x) converge to 0. 3. Mar 13, 2009 ### transgalactic i can split x^2 into xx and i have two sinuses ,one for each x so i do have it the only thing that bothers me is its sin x/x the stuff inside sinus doesnt equal x ?? but they both go to zero ?? 4. Mar 13, 2009 ### sutupidmath Well, it looks rather simple to me, but i might be totally wrong as well. Anyhow, i hope u won't kill me for this :yuck: I can use the fact that both f(x) and g(x) are differentiable at 0...this means that they both are continuous at 0 as well. THis is good to know. NOw i will alsu use the fact that cos(x) is continuous everywhere.this is good to know too. Now: [tex] \lim_{x\rightarrow 0}\frac{cosf(x)-cosg(x)}{x^2}$$

Applying what i just said above, we see that we get a limit of the form 0/0.

So, applying l'hopitals rule twice, will take you home safe.

Edit: DO you know anything about f'(0) and g'(0)??

5. Mar 13, 2009

### transgalactic

nope

you are correct
thanks:)