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. 

Here is how you approach the problem. Put $x=\frac{1}{t}$ and then find the Laurent series at the point $t=0$
$$ f(x) = \frac {(1/t)^3}{(1/t+1)^2}=\frac{1}{t(1+t)^2} ={t}^{-1}-2+3\,t-4\,{t}^{2}+5\,{t}^{3}+ ( {t}^{4}) .$$
Now, substitute $t=1/x$ in the above expansion. See related problem. 
Note: Note that, you only need to find the Taylor series of $\frac{1}{(1+t)^2}$ at $t=0$, since
$$ \frac{1}{t(1+t)^2}= \frac{1}{t} \frac{1}{(1+t)^2} .$$
