## differantiation question ..

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??
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Hi transgalactic! Congrats on your 1001st post!
 Quote by transgalactic f(x) and g(x) are differentiable on 0 f(0)=g(0)=0 calculate $$\lim _{x->0}\frac{cosf(x)-cosg(x)}{x^2}$$
Hint: use L'Hôpital's rule … twice!
 Recognitions: Gold Member Science Advisor Staff Emeritus tiny-tim, that was my first thought. But we don't know that f and g are twice differentiable.

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## differantiation question ..

 Quote by HallsofIvy tiny-tim, that was my first thought. But we don't know that f and g are twice differentiable.
oh tush, mr glass-half-empty …

we don't know that they aren't!

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

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

however did you get that?
 $$\lim _{x->0}\frac{cosf(x)-cosg(x)}{x^2} =\lim _{x->0}\frac{-f'(x)sin(f(x))-f'(x)sing(x)}{2x}=\lim _{x->0}\frac{-f''(x)sin(f(x)+f'(x)cos(f(x))-f''(x)sing(x)+f'(x)cosg(x)}{2}$$ ??
 Blog Entries: 27 Recognitions: Gold Member Homework Help Science Advisor That's better! And if f(0)=g(0)=0, then that = … ?
 Recognitions: Gold Member Science Advisor Staff Emeritus 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?
 its said that" they are differentiable on 0 but not necessarily differentiable around 0

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 Quote by 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?
 Quote by transgalactic its said that" they are differentiable on 0 but not necessarily differentiable around 0
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 )
 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 $$\lim _{x->0}\frac{-f''(x)sin(0)+(f'(x))^2cos(0)-f''(x)sin0+(f'(x))^2cos0}{2}=\lim _{x->0}\frac{(f'(x))^2+(f'(x))^2}{2}$$ ??

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 Quote by 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?
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.
 what is the link between the ability of doing lhopital law and differentiability of a function?
 Recognitions: Homework Help Science Advisor 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'.

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