# differantiation question ..

by transgalactic
Tags: differantiation
 P: 1,399 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!
 PF Patron Sci Advisor Thanks Emeritus P: 38,416 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!

happy days are here again
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happy days are here again!
 P: 1,399 $$\lim _{x->0}\frac{-cosf(x)+cosg(x)}{2}$$ so the solution is 0 ??
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P: 25,496
 Quote by transgalactic $$\lim _{x->0}\frac{-cosf(x)+cosg(x)}{2}$$ so the solution is 0 ??
uhh?

however did you get that?
 P: 1,399 $$\lim _{x->0}\frac{cosf(x)-cosg(x)}{x^2} =\lim _{x->0}\frac{-f'(x)sin(f(x))-f'(x)sing(x)}{2x}=\lim _{x->0}\frac{-f''(x)sin(f(x)+f'(x)cos(f(x))-f''(x)sing(x)+f'(x)cosg(x)}{2}$$ ??
 PF Patron HW Helper Sci Advisor Thanks P: 25,496 That's better! And if f(0)=g(0)=0, then that = … ?
 PF Patron Sci Advisor Thanks Emeritus P: 38,416 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?
 P: 1,399 its said that" they are differentiable on 0 but not necessarily differentiable around 0
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P: 25,496
 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 )
 P: 1,399 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.
 P: 1,399 what is the link between the ability of doing lhopital law and differentiability of a function?
 HW Helper Sci Advisor Thanks P: 24,456 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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P: 25,496
 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 )
 P: 1,399 fixed it $$\lim _{x->0}\frac{-f''(x)sin(0)+(f'(x))^2cos(0)-g''(x)sin0+(g'(x))^2cos0}{2}=\lim _{x->0}\frac{(f'(x))^2+(g'(x))^2}{2}$$ ??
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 Quote by transgalactic fixed it $$\lim _{x->0}\frac{-f''(x)sin(0)+(f'(x))^2cos(0)-g''(x)sin0+(g'(x))^2cos0}{2}=\lim _{x->0}\frac{(f'(x))^2+(g'(x))^2}{2}$$ ??