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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?2012-09-06

4 Answers 4

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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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    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) ).2012-09-06
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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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    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.2012-09-06
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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. 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 you have $\frac{1}{3} x^2 > -ax$ and $\frac{1}{3} x^2 > b$.