1. ## 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?

2. Use l'Hôpital's rule...

3. 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.

4. 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!

5. 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.

6. 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?

7. Originally Posted by HallsofIvy
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.

8. 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!

9. Originally Posted by Showcase_22
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 $\displaystyle \lim \frac{f(x)}{x^2} = 0$

10. 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.