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. 

With equivalents:
At $\infty$, a polynomial is equivalent to its highest degree term, hence a rational function (quotient of polynomials) is equivalent to the  the ratio of the highest degree terms of its numerator and its denominator:
$$\frac{(2n+2) (2n+1)}{(n+1)^2}\sim_\infty\frac{(2n)^2}{n^2}=4.$$
