<s>[INST] Evalute $\lim_{x\to-\infty} \frac{\sqrt{x^2+4x^4}}{8+x^2}$ Having a hard time with this. So far I have:
$$ \frac{\sqrt{x^2(1+4x^2)}}{8+x^2} = \frac{x\sqrt{1+4x^2}}{8+x^2}$$
 [/INST] This is a match-in-heaven for high-school level Non-Standard Analysis (see Keisler textbook).
Since $x$  goes to infinity, we can replace it by the "larger than any real number" $H = \frac {1}{\epsilon} $ where $\epsilon$ is the infinitesimal (smallest than any real number), and dispense with the $\lim$ altogether.
So we can write 
$$\lim_{x\to -\infty}\frac{\sqrt{x^2(1+4x^2)}}{8+x^2} = \lim_{x\to -\infty} \frac{x\sqrt{1+4x^2}}{8+x^2} = \frac {H\sqrt{1+4H^2}}{8+H^2}$$
Note that we continue raising $H$ to powers as though it was a number. To by-pass the square-root, consider the square root of the expression squared:
$$\sqrt {\left(\frac {H\sqrt{1+4H^2}}{8+H^2}\right)^2} = \sqrt {\frac {H^2(1+4H^2)}{64+16H^2 + H^4}} = \sqrt {\frac {H^2+4H^4}{64+16H^2 + H^4}}$$
Take out $H^4$ as common factor,
$$= \sqrt {\frac {H^4(\frac {1}{H^2}+4)}{H^4(\frac {64}{H^4}+\frac {16}{H^2} + 1)}}$$
The term $H^4$ simplifies, and the terms $\frac {1}{H^2}$, $\frac {64}{H^4}$, $\frac {16}{H^2}$ all equal zero. So we ar left with
$$= \sqrt {\frac 41} = 2$$
Who said that you cannot treat infinity as a number? Abraham Robinson formalized it once and for ... the time being.
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