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Differantiation question ..

by transgalactic
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transgalactic
#1
Feb20-09, 03:15 AM
P: 1,397
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??
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tiny-tim
#2
Feb20-09, 06:30 AM
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Hi transgalactic! Congrats on your 1001st post!
Quote Quote by transgalactic View Post
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]
Hint: use L'Hôpital's rule … twice!
HallsofIvy
#3
Feb20-09, 08:27 AM
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tiny-tim, that was my first thought. But we don't know that f and g are twice differentiable.

tiny-tim
#4
Feb20-09, 08:50 AM
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Wink Differantiation question ..

Quote Quote by HallsofIvy View Post
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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transgalactic
#5
Feb20-09, 10:55 AM
P: 1,397
[tex]
\lim _{x->0}\frac{-cosf(x)+cosg(x)}{2}
[/tex]
so the solution is 0
??
tiny-tim
#6
Feb20-09, 11:03 AM
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Quote Quote by transgalactic View Post
[tex]
\lim _{x->0}\frac{-cosf(x)+cosg(x)}{2}
[/tex]
so the solution is 0
??
uhh?

however did you get that?
always show your full calculations!
transgalactic
#7
Feb20-09, 12:06 PM
P: 1,397
[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]
??
tiny-tim
#8
Feb20-09, 12:29 PM
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That's better!

And if f(0)=g(0)=0, then that = … ?
HallsofIvy
#9
Feb20-09, 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?
transgalactic
#10
Feb20-09, 12:45 PM
P: 1,397
its said that" they are differentiable on 0

but not necessarily differentiable around 0
tiny-tim
#11
Feb20-09, 01:29 PM
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Quote Quote by HallsofIvy View Post
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 Quote by transgalactic View Post
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 )
transgalactic
#12
Feb20-09, 02:34 PM
P: 1,397
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]

??
Dick
#13
Feb20-09, 02:39 PM
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Quote Quote by HallsofIvy View Post
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.
transgalactic
#14
Feb20-09, 03:05 PM
P: 1,397
what is the link between the ability of doing lhopital law
and differentiability of a function?
Dick
#15
Feb20-09, 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'.
tiny-tim
#16
Feb20-09, 04:26 PM
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Quote Quote by transgalactic View Post
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]

??
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 )
transgalactic
#17
Feb21-09, 01:26 AM
P: 1,397
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]

??
tiny-tim
#18
Feb21-09, 03:14 AM
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Quote Quote by transgalactic View Post
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]

??
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?


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