# Differantiation question

1. Feb 20, 2009

### transgalactic

f(x) and g(x) are differentiable on 0
f(0)=g(0)=0

calculate
$$\lim _{x->0}\frac{cosf(x)-cosg(x)}{x^2}$$
??

it looks like cauchys mean value theorem
F(x)=cos(f(x))
G(x)=cos(g(x))
what to do next??

2. Feb 20, 2009

### tiny-tim

Welcome to the second millenium!

Hi transgalactic! Congrats on your 1001st post!
Hint: use L'Hôpital's rule … twice!

3. Feb 20, 2009

### HallsofIvy

tiny-tim, that was my first thought. But we don't know that f and g are twice differentiable.

4. Feb 20, 2009

### tiny-tim

always look on the bright side of life … ♩ ♫ ♪ ♫ ♬ ♩

oh tush, mr glass-half-empty …

we don't know that they aren't!

happy days are here again
the skies above are clear again
let us sing a song of cheer again
happy days are here again!

5. Feb 20, 2009

### transgalactic

$$\lim _{x->0}\frac{-cosf(x)+cosg(x)}{2}$$
so the solution is 0
??

6. Feb 20, 2009

### tiny-tim

uhh?

however did you get that?

7. Feb 20, 2009

### transgalactic

$$\lim _{x->0}\frac{cosf(x)-cosg(x)}{x^2} =\lim _{x->0}\frac{-f'(x)sin(f(x))-f'(x)sing(x)}{2x}=\lim _{x->0}\frac{-f''(x)sin(f(x)+f'(x)cos(f(x))-f''(x)sing(x)+f'(x)cosg(x)}{2}$$
??

8. Feb 20, 2009

### tiny-tim

That's better!

And if f(0)=g(0)=0, then that = … ?

9. Feb 20, 2009

### HallsofIvy

Transgalactic, as much as I hate to pour cold water on the party, you started by saying "if f and g are differentiable at 0". Is that all you are given or are you given that the second derivative exists?

10. Feb 20, 2009

### transgalactic

its said that" they are differentiable on 0

but not necessarily differentiable around 0

11. Feb 20, 2009

### tiny-tim

This is a bit of a technicality … we'll discuss why in a moment …

but first, just answer my previous question: what is that limit equal to at x = 0?

(oh, btw, just noticed … you should have had an f'(x)2 somewhere in that )

12. Feb 20, 2009

### transgalactic

i fixed it to
$$\lim _{x->0}\frac{cosf(x)-cosg(x)}{x^2} =\lim _{x->0}\frac{-f'(x)sin(f(x))-f'(x)sing(x)}{2x}=\lim _{x->0}\frac{-f''(x)sin(f(x)+(f'(x))^2cos(f(x))-f''(x)sing(x)+(f'(x))^2cosg(x)}{2}$$

after i did lhopital twice i got
$$\lim _{x->0}\frac{-f''(x)sin(0)+(f'(x))^2cos(0)-f''(x)sin0+(f'(x))^2cos0}{2}=\lim _{x->0}\frac{(f'(x))^2+(f'(x))^2}{2}$$

??

13. Feb 20, 2009

### Dick

I'm with Halls. You only know the derivatives exist AT zero. You can't even apply l'Hopital the first time, you can't take the limit of something you only know exists at a single point. I'd suggest using the series expansion for cos(x) and difference quotients.

14. Feb 20, 2009

### transgalactic

what is the link between the ability of doing lhopital law
and differentiability of a function?

15. Feb 20, 2009

### Dick

Take f(x)=x^2*sin(1/x^2), f(x)=0. That's differentiable at 0. In fact, it's differentiable everywhere. But it's not continuously differentiable at 0. You can't do anything with l'Hopital at 0. But that doesn't mean your limit doesn't exist. So the problem with l'Hopital is more than 'technical'.

16. Feb 20, 2009

### tiny-tim

oh, transgalactic, however do you manage to make these mistakes?

you started with an expression that was half f and half g,

but you ended losing all the g

try again

(and then we'll sort out the technicalitites )

17. Feb 21, 2009

### transgalactic

fixed it
$$\lim _{x->0}\frac{-f''(x)sin(0)+(f'(x))^2cos(0)-g''(x)sin0+(g'(x))^2cos0}{2}=\lim _{x->0}\frac{(f'(x))^2+(g'(x))^2}{2}$$

??

Last edited: Feb 21, 2009
18. Feb 21, 2009

### tiny-tim

much better!

(though you still need to fiddle around with those pluses and minuses )

EDIT: oh, and since f'(0) and g'(0) are both defined as existing, is there any reason why you can't use them at the end?

Last edited: Feb 21, 2009
19. Feb 21, 2009

### transgalactic

fixed it
$$\lim _{x->0}\frac{-f''(x)sin(0)-(f'(x))^2cos(0)+g''(x)sin0+(g'(x))^2cos0}{2}=\lim _{x->0}\frac{-(f'(x))^2+(g'(x))^2}{2}$$

??

Last edited: Feb 21, 2009
20. Feb 21, 2009

what now??