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

Often "limit does not exist" means that the limit is also not $+\infty$ or $-\infty$, e.g. the sequence oscillates forever like $\sin n$. In your case, one could say the limit exists but it is $+\infty$. To see this, you can note that $n^3 + 2 > (n/4)(n^2 + 3)$ when $n \geq 1$, because $n^2 + 3 \leq 4n^2$. So your fraction is always greater than $n/4$, which goes to $+\infty$.  
