
#1
Feb2009, 03:15 AM

P: 1,399

f(x) and g(x) are differentiable on 0
f(0)=g(0)=0 calculate [tex] \lim _{x>0}\frac{cosf(x)cosg(x)}{x^2} [/tex] ?? it looks like cauchys mean value theorem F(x)=cos(f(x)) G(x)=cos(g(x)) what to do next?? 



#2
Feb2009, 06:30 AM

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Hi transgalactic! Congrats on your 1001st post!




#3
Feb2009, 08:27 AM

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




#4
Feb2009, 08:50 AM

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differantiation question .. 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
Feb2009, 10:55 AM

P: 1,399

[tex]
\lim _{x>0}\frac{cosf(x)+cosg(x)}{2} [/tex] so the solution is 0 ?? 



#6
Feb2009, 11:03 AM

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however did you get that? always show your full calculations! 



#7
Feb2009, 12:06 PM

P: 1,399

[tex]
\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} [/tex] ?? 



#8
Feb2009, 12:29 PM

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That's better!
And if f(0)=g(0)=0, then that = … ? 



#9
Feb2009, 12:40 PM

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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
Feb2009, 12:45 PM

P: 1,399

its said that" they are differentiable on 0
but not necessarily differentiable around 0 



#11
Feb2009, 01:29 PM

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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
Feb2009, 02:34 PM

P: 1,399

i fixed it to
[tex] \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} [/tex] after i did lhopital twice i got [tex] \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} [/tex] ?? 



#13
Feb2009, 02:39 PM

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#14
Feb2009, 03:05 PM

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



#15
Feb2009, 03:12 PM

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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
Feb2009, 04:26 PM

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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
Feb2109, 01:26 AM

P: 1,399

fixed it
[tex] \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} [/tex] ?? 


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