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We want to prove $\lim_{x\to +\infty}\frac{x^3}{|x|} = \infty$. I don't think the usual "$\varepsilon-\delta$ definition would work since we are dealing with $\infty$. How else could I approach this problem?

I know plugging in $\infty$ would show this. But is there a more rigorous proof?

For example, if I wanted to change this problem so that $x\to0$ the plugging-in method would not work anymore since we cannot divide by $0$. So there must be a way to solve this so that this problem does not arise!

Thank you.

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  • $\begingroup$ Depends what you mean by "the usual". The definition of$$f(x)\to\infty\quad\hbox{as}\quad x\to\infty$$is$$\forall M\in{\Bbb R}\ \exists X\in{\Bbb R}\ \forall x\in{\Bbb R}((x>X)\Rightarrow(f(x)>M))\ .$$ $\endgroup$
    – David
    May 22, 2015 at 2:24
  • $\begingroup$ Well you can't say any value "=" $\infty$. I'm not sure what the right notation is to use because I have long argued that divergent limits are quite a bit different than simply indeterminate limits (e.g. oscillating limits). $\endgroup$
    – Jared
    May 22, 2015 at 2:27
  • $\begingroup$ So how do you deal with the case when we are finding $lim_{x\to \infty} x$? This clearly grows infinitely large, so if we cannot say this $lim=\infty$, what do we say? $\endgroup$ May 22, 2015 at 2:29
  • $\begingroup$ BTW the "$\epsilon$-$\delta$" definition doesn't work because this limit doesn't exist (which can be showed by using an $\epsilon$-$\delta$ proof). $\endgroup$
    – Jared
    May 22, 2015 at 2:30
  • $\begingroup$ This is true! Thank you. That's the perfect way to articulate what I was thinking :P. $\endgroup$ May 22, 2015 at 2:31

3 Answers 3

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For positive $x$, $|x|=x$, so

$$\lim\limits_{x\to\infty} \frac{x^3}{|x|}=\lim\limits_{x\to\infty} \frac{x^3}{x}=\lim\limits_{x\to\infty} x^2=\infty$$

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  • $\begingroup$ Thank you! You've answered another one of my questions before too, haha. $\endgroup$ May 22, 2015 at 2:30
  • $\begingroup$ @DavidSouth of course :) $\endgroup$
    – user223391
    May 22, 2015 at 2:31
  • $\begingroup$ What would you do in the case where we are taking the limit as $x \to -\infty$? We have negative $x$ in this case. Can we still use the logic from your answer? $\endgroup$ May 22, 2015 at 2:46
  • $\begingroup$ Just note that for negative $x$, $|x|=-x$ and the answer will be $-\infty$. $\endgroup$
    – user223391
    May 22, 2015 at 3:03
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Note that for $x > 0$, $|x| = x$. Since we're taking $x$ to infinity, $x$ must pass $0$ at some point, so we can claim

$$\lim_{x\to +\infty} \frac{x^3}{|x|} = \lim_{x \to +\infty} \frac{x^3}{x}.$$

From here you can make two arguments: either we discount the removable singularity at $x=0$, because we're onwards to $+\infty$ anyways, or we can use L'Hopital. In either case, we'll get something that is proportional to $\lim_{x\to+\infty} x^2$.

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  • $\begingroup$ Thank you! How about in the situation where $x\to-\infty$? What would we do with the absolute value in that situation then? $\endgroup$ May 22, 2015 at 2:27
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    $\begingroup$ Well, what is $|x|$ when $x < 0$? $\endgroup$
    – Emily
    May 22, 2015 at 2:27
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An application of L'Hopital's rule can be used here, since we have something of the form $\frac{\infty}{\infty}$. This gives \begin{equation*} \lim_{x\to\infty}\frac{3x^2}{|x|'} \end{equation*} where $'$ denotes the derivative with respect to $x$. Since the denominator has a horizontal asymptote at $1$, we have \begin{equation*} 3\lim_{x\to\infty}x^2=\infty. \end{equation*}

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