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L'hopital's case proof, infinite limit

  • #1

Homework Statement


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

[tex]g(x), g'(x)[/tex] not equal to zero

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

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

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 [tex]N[/tex] there is a [tex]\delta > 0[/tex] s.t. [tex]f'(x)/g'(x) > N[/tex] when [tex]0 < |x - a| < \delta[/tex]
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?
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
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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.
 
  • #3
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
 
  • #4
Actually, I think I can use [tex] |f'(x)/g'(x)| > 1/ \epsilon[/tex], but that still doesn't seem too much more helpful
 
Last edited:
  • #5
Dick
Science Advisor
Homework Helper
26,258
618
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.
 
  • #6
Oh man, okay. I will update if I get this
 

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