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Thread: True or false?

  1. #1
    Super Member Showcase_22's Avatar
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    True or false?

    Suppose that $\displaystyle f0, \infty) \rightarrow \mathbb{R}$ is twice differentiable and $\displaystyle \lim_{x \rightarrow \infty} f''(x)=0$.

    Is it true that $\displaystyle \lim_{x \rightarrow \infty} \frac{f(x)}{x^2}=0$?
    I've been rooting around for a counterexample in vain so I think this is true.

    However, I have no idea how to prove it is true!

    Does anyone have any ideas?
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  2. #2
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    Use l'Hôpital's rule...
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  3. #3
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    Good idea, but

    Since $\displaystyle \lim_{x\rightarrow \infty} x^2= \infty$, L'Hopital's rule only applies if $\displaystyle \lim_{x\rightarrow \infty} f(x)= \infty$ also. That would have to be shown.
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  4. #4
    Super Member Showcase_22's Avatar
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    I've already tried using L'Hopital's rule and it got me nowhere.

    Here's what I did:

    $\displaystyle

    \lim_{x \rightarrow \infty} \frac{f(x)}{x^2}
    \Rightarrow \ \frac{\lim_{x \rightarrow \infty} f(x)}{\lim_{x \rightarrow \infty} x^2}$

    Suppose this tends to a limit a where $\displaystyle a \neq 0$.

    $\displaystyle \frac{\lim_{x \rightarrow \infty} f(x)}{\lim_{x \rightarrow \infty} x^2}=a \Rightarrow \lim_{x \rightarrow \infty} f(x)= \infty $

    So L' Hopital's rule applies.

    Therefore $\displaystyle \frac{\lim_{x \rightarrow \infty} f(x)}{\lim_{x \rightarrow \infty} x^2}=\frac{\lim_{x \rightarrow \infty} f'(x)}{\lim_{x \rightarrow \infty} 2x}=\frac{\lim_{x \rightarrow \infty} f''(x)}{\lim_{x \rightarrow \infty} 2}=0$ since $\displaystyle \lim_{x \rightarrow \infty} f''(x)=0.$

    However this contradicts our definition of a!
    Last edited by Showcase_22; Apr 30th 2009 at 11:03 AM.
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  5. #5
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    You defined a that way:
    $\displaystyle a = \lim_{x \rightarrow \infty} \frac{f(x)}{x^2}$. And our point was to prove $\displaystyle a = 0$, and it was done by L'Hopital's rule.
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  6. #6
    Super Member Showcase_22's Avatar
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    but if a=0 then the step where $\displaystyle

    \frac{\lim_{x \rightarrow \infty} f(x)}{\lim_{x \rightarrow \infty} x^2}=a \Rightarrow \lim_{x \rightarrow \infty} f(x)= \infty
    $ fails.

    I thought this made the proof incorrect?
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  7. #7
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    Quote Originally Posted by HallsofIvy View Post
    Good idea, but

    Since $\displaystyle \lim_{x\rightarrow \infty} x^2= \infty$, L'Hopital's rule only applies if $\displaystyle \lim_{x\rightarrow \infty} f(x)= \infty$ also. That would have to be shown.
    Of course when f is bounded it is trivial.
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  8. #8
    Super Member Showcase_22's Avatar
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    Of course when f is bounded it is trivial.
    Is it?

    If f is bounded then $\displaystyle \lim_{x \rightarrow \infty} f(x)< \infty$ or it doesn't exist.

    We want $\displaystyle \lim_{ x \rightarrow \infty} f(x)= \infty$, we don't want it to converge or not exist!
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  9. #9
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    Quote Originally Posted by Showcase_22 View Post
    but if a=0 then the step where $\displaystyle

    \frac{\lim_{n \rightarrow \infty} f(x)}{\lim_{n \rightarrow \infty} x^2}=a \Rightarrow \lim_{n \rightarrow \infty} f(x)= \infty
    $ fails.

    I thought this made the proof incorrect?
    You did not. Limit of quotients is quotient of limits when the limits exists (and are not infinite).

    I think the point was, what happens when $\displaystyle \lim_{x\rightarrow \infty}f(x) \neq \infty$. It can happen when f is bounded. We have then $\displaystyle -\frac{M}{x^2} < \frac{f(x)}{x^2} < \frac{M}{x^2}$. And hence the limit $\displaystyle \lim \frac{f(x)}{x^2} = 0$
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  10. #10
    Super Member Showcase_22's Avatar
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    ohh, now I getcha!!

    So my proof should go like this:

    Assume $\displaystyle \lim_{x \rightarrow \infty} f(x)= \infty$. This means that L' Hopital's rule can be applied:

    $\displaystyle

    \lim_{x \rightarrow \infty} \frac{f(x)}{x^2}
    \Rightarrow \ \frac{\lim_{x \rightarrow \infty} f(x)}{\lim_{x \rightarrow \infty} x^2}
    $$\displaystyle

    \Rightarrow \frac{\lim_{x \rightarrow \infty} f(x)}{\lim_{x \rightarrow \infty} x^2}=\frac{\lim_{x \rightarrow \infty} f'(x)}{\lim_{x \rightarrow \infty} 2x}=\frac{\lim_{x \rightarrow \infty} f''(x)}{\lim_{x \rightarrow \infty} 2}=0
    $ since $\displaystyle

    \lim_{x \rightarrow \infty} f''(x)=0.
    $.

    The first case yields $\displaystyle \lim_{x \rightarrow \infty} \frac{f(x)}{x^2}=0$.

    Now assume that $\displaystyle \lim_{ x \rightarrow \infty} \frac{f(x)}{x^2} \neq \infty \Rightarrow \ f(x)$ is bounded.

    Let $\displaystyle f(x)$ be bounded by $\displaystyle M \in \mathbb{R}$ $\displaystyle \Rightarrow \ -M \leq f(x) \leq M \Rightarrow \ -\frac{M}{x^2} \leq \frac{f(x)}{x^2} \leq \frac{M}{x^2} \Rightarrow \ \lim_{x \rightarrow \infty} \frac{f(x)}{x^2}=0$.

    Hence in all cases $\displaystyle \lim_{x \rightarrow \infty} \frac{f(x)}{x^2}=0$ so the statement is true.
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