# Suppose$\lim \limits_{n \to \infty}a_n$=A; $\lim \limits_{n \to \infty}b_n$=B. Use limit laws to find limit of $\frac{2a_n+3nb_n}{4n+1}$.

My only guess is to start with $\frac{2a_n+3nb_n}{4n+1}$ = $\frac{2a_n}{4n+1}+\frac{3nb_n}{4n+1}$=$[\frac{2}{4n+1}*(a_n)]$+[$\frac{3n}{4n+1}*(b_n)$] and then I don't really know where to go from there. I'm not even sure if that's going in the right direction.

edit: Okay, so since we know that $\frac{2}{4n+1}$ converges to 0 and $\frac{3n}{4n+1}$ converges to $\frac{3}{4}$, and we know that the product of two convergent sequences is convergent...could we split all of them up and get that $\lim \limits_{n \to \infty}\frac{2}{4n+1}$$\lim \limits_{n \to \infty}a_n + \lim \limits_{n \to \infty}\frac{3n}{4n+1}$$\lim \limits_{n \to \infty}b_n$ = 0*A + $\frac{3}{4}*B$ = $\frac{3B}{4}$? Am I using the limit laws correctly?

Starting off with $\frac{2}{4n+1}*(a_n) + \frac{3n}{4n+1}*(b_n)$. Let $c_n = \frac{2}{4n+1}$ and $d_n = \frac{3n}{4n+1}$. Notice that $c_n \rightarrow 0$ and $d_n \rightarrow 3/4$. Then by the limit laws $\lim a_n * c_n + b_n * d_n = 0*A + 3/4*B = 3/4 * B$.