Formally prove  $ \lim_{x \rightarrow -\infty} x^2+30x-1000= \infty$ 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.
 A: 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 $x^2+30x-1000 > L$.
A: 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 $\lim_{x\rightarrow-\infty}x^2+30x-1000=+\infty$.
A: $$x^2+30x-1000=(x+50)(x-20)>(x+50)x=x^2+50x\geq 100x\,\,,\,\,for\,\,\,x>2\Longrightarrow$$
$$\Longrightarrow\forall\,R\in\Bbb R^+\,\,,\,\,take\,\,\,x>\frac{R}{100}$$
A: 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 $\frac{1}{3} x^2 > b$.  
