<s>[INST] Finding limits of rational functions as $x\to\infty$ 
Possible Duplicate:
Finding the limit of $\frac{Q(n)}{P(n)}$ where $Q,P$ are polynomials 


$$\displaystyle \lim_{x\to\infty}\frac{(2x^2+1)^2}{(x-1)^2(x^2+x)}.$$

I do not know where to start on this, I tried multiplying it out and that didn't help really. It seems very complicated and I know I have to reduce it somehow but everything I do just makes it more complicated.
 [/INST] By substitution. 
I think the OP is having some difficulty in dividing by a power of $x$ and simplifying the rational function, in terms of $1/x$. I must admit that the simplification process seems a bit unnatural while working with $1/x$.
So, my suggestion is that, at the very beginning, substitute $y = 1/x$. Then $x \to \infty$ is equivalent to $y \to 0$. (Note that we should be talking about the one-sided limit as $x \to 0^+$, but we will ignore that issue here.) 
Plugging this in the numerator, we get:
$$
(2x^2+1)^2 = \left(2\frac{1}{y^2} + 1\right)^2 = \left(\frac{2+y^2}{y^2}\right)^2 = \frac{(2+y^2)^2}{y^4}.
$$
Similarly the denominator becomes:
$$
(x-1)^2(x^2+x) = \left( \frac{1}{y} - 1\right)^2 \left( \frac{1}{y^2} + \frac{1}{y} \right) = \frac{(1-y)^2}{y^2} \times \frac{1+y}{y^2} = \frac{(1-y)^2(1+y)}{y^4}.
$$
Dividing the numerator by the denominator, we get
$$
\frac{(2x^2+1)^2}{(x-1)^2(x^2+x)} = \frac{(2+y^2)^2}{y^4} \times \frac{y^4}{(1-y)^2(1+y)} 
=  \frac{(2+y^2)^2}{(1-y)^2(1+y)}.
$$
(Notice that the problematic $y^4$ terms nicely cancel.)
Finally, we have
$$
\lim_{x \to \infty} \frac{(2x^2+1)^2}{(x-1)^2(x^2+x)} 
= 
\lim_{y \to 0} \frac{(2+y^2)^2}{(1-y)^2(1+y)}.
$$
I believe that the latter limit should be easy to solve (essentially by plugging in $0$).
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