differantiation question ..

transgalactic

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??

tiny-tim

Hi transgalactic! Congrats on your 1001st post!
Hint: use L'Hôpital's rule … twice!

HallsofIvy

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

tiny-tim

oh tush, mr glass-half-empty …

transgalactic

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

tiny-tim

uhh?

however did you get that?

always show your full calculations!

transgalactic

[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

That's better!

And if f(0)=g(0)=0, then that = … ?

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?

transgalactic

its said that" they are differentiable on 0

but not necessarily differentiable around 0

tiny-tim

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

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

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

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

Dick

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

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

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]

??