L'hopital's case proof, infinite limit

[Degeneration]

Homework Statement

a in R is finite, f,g are differentiable on R
$$\lim_{\substack{x\rightarrow a}} f(x)=\infty$$
$$\lim_{\substack{x\rightarrow a}} g(x)=\infty$$

$$g(x), g'(x)$$ not equal to zero

$$\lim_{\substack{x\rightarrow a}} f'(x)/g'(x)=\infty$$

Show $$\lim_{\substack{x\rightarrow a}} f(x)/g(x)=\infty$$

Homework Equations

I'm sure you need to use the MVT
f'(c)/g'(c) = (f(x) - f(a))/(g(x) - g(a))

The Attempt at a Solution

I'm starting out trying to use the continuity definition, but it seems to be going nowhere with a infinite limit.

For every number $$N$$ there is a $$\delta > 0$$ s.t. $$f'(x)/g'(x) > N$$ when $$0 < |x - a| < \delta$$
Additionally, I can't just say lim x->a f'(x)/g'(x) = infinity = L and then use epsilon delta, since I don't know if it works for extended reals. Where can I go from here?

 

[Dick]

You know for every N there is an 'a' such that f'(t)/g'(t)>N for all x>a. Write that as f'(t)>N*g'(t) and integrate both sides from a to x. Then divide by g(x) and think about the limit as x->infinity.

 

[Degeneration]

Right, sorry though I forgot to mention this is strictly differentiation. No Riemann sums or integrals are allowed. I understand how that would make it significantly easier though

 

[Degeneration]

Actually, I think I can use $$|f'(x)/g'(x)| > 1/ \epsilon$$, but that still doesn't seem too much more helpful

 

[Dick]

Right, sorry though I forgot to mention this is strictly differentiation. No Riemann sums or integrals are allowed. I understand how that would make it significantly easier though

Hmm. I guess I'm not really seeing how to do it then. The trouble with your MVT statement is that you know (f(x)-f(a))/(x-a)=f'(c) and (g(x)-g(a))/(x-a)=g'(d), for some values c and d in [a,x], but you don't know that c=d.

 

[Degeneration]

Oh man, okay. I will update if I get this