<s>[INST] Taylor's series when x goes to infinity Let $f(x) = \frac {x^3}{(x+1)^2}$. Find constants a, b, c, so that $f(x) = ax + b + \frac cx +  o(\frac 1x)$ as $x$ goes to $\pm \infty$. So i know that i can't take Taylor series as $x$ goes to infinity. So i am assuming i have to make some kind of substitution. I tried making $x = \frac 1u$ but i get nowhere.
$$f(x) = \frac {x^3}{(x+1)^2} = x^3(x+1)^{-2}=\frac{1}{u^3}(1 + \frac 1u)$$ 
$$\lim_{u\to 0}\frac{1}{u^3}(1 + \frac 1u) = \lim_{u\to 0}\frac{1}{u^3}(1 - \frac 2u  + \frac{3}{u^2} + o({u^3})) =\lim_{u\to 0} u^3 - \frac {2}{u^4} + \frac{3}{u^5} + o(\frac 1u)    = \lim_{u\to 0} \frac{1}{x^3} - 2x^4 + 3x^5 + o(x^5)$$
Thanks. 
 [/INST] I would start off here by using long division to simplify things a bit:
$$
\frac{x^3}{(x+1)^2}=\frac{x^3}{x^2+2x+1}=\cdots=x-2+\frac{3x+2}{x^2+2x+1}.
$$
Heuristically speaking, it is clear that as $x\rightarrow\infty$, that last bit is $\frac{3}{x}+o(\frac{1}{x})$.  So, this would suggested that
$$
\frac{x^3}{(x+1)^2}=x-2+\frac{3}{x}+o\left(\frac{1}{x}\right)\text{ as }x\to\infty.
$$
Having used long division / heuristics to find the answer, now it is only a matter of justifying it; and, that shouldn't be too bad.
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