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??
Welcome to the second millenium! Hi transgalactic! Congrats on your 1001st post! Hint: use L'Hôpital's rule … twice!
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[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] ??
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?
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 )
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] ??
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.
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'.
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 )
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?
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] ??