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calculus TA here, trying to come up with good outside-the-box questions for my students, this one turns out to be subtler and harder than meets the eye-- I can't find a solution which satisfies me, one which isn't too ad hoc!

The problem is: prove, using the formal definition of limits, that $$\lim_{x\to -\infty} x^2+30x-1000=\infty$$

Of course, the $-1000$ is just a red herring. But the 30x term seems to nontrivially complicate the problem a bit.

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What about $x^2 + 30x = (x + 15)^2 - 15^2$? Is this useful? –  Tunococ Sep 6 '12 at 4:57
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4 Answers

If $x \leq -31$, then $x^2+30x-1000 = x(x+30)-1000 \geq -x-1000$.

Choose $L>0$. Then if $x < -(L+1000)$, you can easily check that $x^2+30x-1000 > L$.

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Thanks. Here we implicitly are using the fact that if $x<-(L+1000)$ then $x\leq -31$ (an assumption used in the first line). This wouldn't work if the 1000 were replaced by, say, 1 (or rather, it would work but the 1000 would be pulled out of thin air). Aside from that, what's written on your scratch paper to let you deduce the above? It's probably more important to show them the thought process. Thanks again for the quick answer! –  Anonymous Sep 6 '12 at 5:25
    
The only other line on my scratch paper is $-x-1000 > L$ from which I conclude that I need $x < -(L+1000)$. If the number was $c$ instead of $1000$, I would have used $x < \min ( -31, -(L+c) )$. –  copper.hat Sep 6 '12 at 6:03
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$$x^2+30x-1000=(x+50)(x-20)>(x+50)x=x^2+50x\geq 100x\,\,,\,\,for\,\,\,x>2\Longrightarrow$$

$$\Longrightarrow\forall\,R\in\Bbb R^+\,\,,\,\,take\,\,\,x>\frac{R}{100}$$

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See, this proves the problem's subtlety. $x$ is not approaching $\infty$, it is approaching $-\infty$, so taking $x>R/100$ is not a good move. Thanks for pointing out that the original function factors nicely-- I didn't realize that at all and certainly didn't intend it, I will change the 1000 to a 999 just to avoid that when I do this problem in class! :) –  Anonymous Sep 6 '12 at 5:27
    
Well, perhaps after the editing it was clearer that it was not $\,x\to\infty\,$ but $\,x\to -\infty\,$ , but if something helped anyway then good. –  DonAntonio Sep 6 '12 at 8:16
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Clearly the $x^2$ term is all that matters; to see this, write $x^2+30x-1000=x^2\left(1+\frac{30}{x}-\frac{1000}{x^2}\right)$, and restrict attention to $\lvert{x}\rvert>100$, in which case $\lvert{30/x}\rvert<0.3$ and $\lvert{1000/x^2}\rvert < 0.1$. The term in parentheses is therefore at least $3/5$. Now for any $L$, let $N=\min(-100,-\sqrt{5/3L})$; we have $x^2+30x-1000 \ge \frac{3}{5}x^2\ge L$ for all $x<N$. Since $L$ was arbitrary, we conclude that $\lim_{x\rightarrow-\infty}x^2+30x-1000=+\infty$.

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You can note that if $x <-90 $ then $\frac{1}{3} x^2 > -30x$ and $\frac{1}{3} x^2 > 1000$. Thus, for all $x <-90$ we have

$$x^2+30x-1000 > \frac{x^2}{3} \,.$$

You can now easily finish the problem either by a standard $\epsilon-\delta$ argument, or, if the students know it, by squeezing it.

P.S. No matter what $a,b$ you chose, you can easely find a $c$ so that for all $x <c $ you have $\frac{1}{3} x^2 > -ax$ and $\frac{1}{3} x^2 > b$.

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