<s>[INST] Different ways to find limit in infinity I would like to find a limit  of the function $f(x)=\frac{x^3+\sqrt{x^6-1}}{4(x-2)^2}$ as $x\rightarrow \infty$ and as $x\rightarrow -\infty$.
 In the first case, both  numerator and denominator goes to infinity but nominator has  higher exponent so $$\lim_{x\rightarrow \infty}\frac{x^3+\sqrt{x^6-1}}{4(x-2)^2}= \infty.$$
In the second case numerator is indefinite $-\infty+\infty$ so I can do "the trick" where  I transform the expression by factoring with $x^3-\sqrt{x^6-1}$ and I get
$$\frac{x^3+\sqrt{x^6-1}}{4(x-2)^2}=\frac{x^6-x^6+1}{4(x-2)^2(x^3-\sqrt{x^6-1})}=\frac{1}{4(x-2)^2(x^3-\sqrt{x^6-1})}$$
which tends to $0$ as $x\rightarrow-\infty$. Is this step correct? It seems wrong since I cant think of the reason why I couldn't do the same in the first case and get limit $0$ as $x\rightarrow \infty$ which would, of course, be wrong. 
What am I missing?
 [/INST] Something I learned when dealing with limits $x \to -\infty$ is that you can do a change in variable.
$$
\lim_{x \to - \infty} \frac{x^3 + \sqrt{x^6 -1}}{4(x-2)^2} \left/ x = -t, \ x \to -\infty \Rightarrow t \to \infty \right/ = \lim_{t \to \infty} \frac{-t^3 + \sqrt{t^6 -1}}{-4(t+2)^2}
$$
Now for big values of $t$ this will behave like
$$
\frac{-t^3 + \sqrt{t^6 -1}}{-4(t+2)^2} \approx \frac{-t^3 + t^3}{-4(t+2)^2} = 0
$$
I hope this illustrates the difference.
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