Limit of differentiable functions question

[transgalactic]

f(x) and g(x) are differentiable on 0
f(0)=g(0)=0
<br /> \lim _{x-&gt;0}\frac{cos(f(x))-cos(g(x))}{x^2}=\lim _{x-&gt;0}\frac{-2sin(\frac{f(x)+g(x)}{2})sin(\frac{f(x)-g(x)}{2})}{x^2}=-2<br />

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

 

[HallsofIvy]

No, you don't have "sin(x)/x", you have
\frac{sin(\frac{f(x)+ g(x)}{2})}{x^2}[/itex]<br /> <br /> whether this converges at all depends on how fast f(x) and g(x) converge to 0.

 

[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 doesn't equal x
??
but they both go to zero
??

 

[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


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:


\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'hospital's rule twice, will take you home safe.

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

 

[transgalactic]

nope

you are correct
thanks:)