<s>[INST] How to solve this limit: $\lim_{n \to \infty} \frac{(2n+2) (2n+1) }{ (n+1)^2}$ $$
\lim_{n\to\infty}\frac{(2n+2)(2n+1)}{(n+1)^{2}}
$$ 
When I expand it gives:
$$
\lim_{n\to\infty} \dfrac{4n^{2} + 6n + 2}{n^{2} + 2n + 1}
$$
How can this equal $4$? Because if I replace $n$ with infinity it goes $\dfrac{\infty}{\infty}$ only. 
 [/INST] To handle limits involving fractions you factor out the dominant term from top and bottom such that they cancel, by dominant term; I mean the term of highest degree, in this case it's $n^2$: $$\lim_{n\to\infty} \frac{4n^2 + 6n + 2}{n^2 + 2n + 1}$$
$$=\require\cancel\lim_{n\to\infty} \frac{2\cancel{n^2}\left(2 + \frac{3}{n} + \frac{1}{n^2}\right)}{\cancel{n^2}\left(1 + \frac{2}{n} + \frac{1}{n^2}\right)}$$
$$=\lim_{n\to\infty} \frac{2\left(2 + \frac{3}{n} + \frac{1}{n^2}\right)}{1 + \frac{2}{n} + \frac{1}{n^2}}\tag{1}$$
$$=\frac{2\left(2 + 0 + 0\right)}{1 + 0 + 0}\tag{2}$$
$$=\color{blue}{4}$$
You get from $(1)$ to $(2)$ by making the observation that each of the fractions with $n$ or $n^2$ in the denominator will equal zero in the limit as $n$ tends to infinity. 
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