Register to reply 
Differantiation question ..by transgalactic
Tags: differantiation 
Share this thread: 
#1
Feb2009, 03:15 AM

P: 1,398

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

Sci Advisor
HW Helper
Thanks
P: 26,160

Hi transgalactic! Congrats on your 1001st post!



#3
Feb2009, 08:27 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,294

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



#4
Feb2009, 08:50 AM

Sci Advisor
HW Helper
Thanks
P: 26,160

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: 1,398

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


#6
Feb2009, 11:03 AM

Sci Advisor
HW Helper
Thanks
P: 26,160

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


#7
Feb2009, 12:06 PM

P: 1,398

[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

Sci Advisor
HW Helper
Thanks
P: 26,160

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


#9
Feb2009, 12:40 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,294

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: 1,398

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


#11
Feb2009, 01:29 PM

Sci Advisor
HW Helper
Thanks
P: 26,160

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: 1,398

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

Sci Advisor
HW Helper
Thanks
P: 25,251




#14
Feb2009, 03:05 PM

P: 1,398

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


#15
Feb2009, 03:12 PM

Sci Advisor
HW Helper
Thanks
P: 25,251

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

Sci Advisor
HW Helper
Thanks
P: 26,160

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: 1,398

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] ?? 


Register to reply 
Related Discussions  
Differantiation proof question ..  Calculus & Beyond Homework  11  
Differantiation proof question ..  Calculus & Beyond Homework  5  
Differantiation proof question exlanation problem..  Calculus & Beyond Homework  2  
Differantiation proof question..  Calculus & Beyond Homework  1  
Differantiation proof question..  Calculus & Beyond Homework  3 