Limit $\lim_{n\to\infty}\frac{n^3+2}{n^2+3}$ Find the limit and prove your answer is correct
$$\lim_{n\to\infty}\frac{n^3+2}{n^2+3}$$
By divide everything by $n^3$ I got  
$$\lim_{n\to\infty}\frac{n^3+2}{n^2+3}=\frac10 $$
which is undefined. So I conclude that there is no limit for this sequence. However, I don't know how to prove it

Hint:
$$\frac{n^3+2}{n^2+3}=\frac{n^3+3n-3n +2 }{n^2+3}= \frac{n(n^2+3)}{n^2+3} +\frac{2-3n}{n^2+3}=n +\frac{2-3n}{n^2+3} $$
