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
 A: Your basic strategy works, but it may be clearer to divide top and bottom by $n^2$. We get
$$\frac{n+\frac{2}{n^2}}{1+\frac{3}{n^2}}.$$
As $n\to\infty$, the bottom approaches $1$, while the top blows up. 
So when $n$ is very large, our function is very large. As $n$ grows without bound, so does our function. 
The final answer now depends on the terminology used in your course. Some people would say that 
$$\lim_{n\to\infty}\frac{n^3+2}{n^2+3}$$
does not exist. (From your post, it looks as if that is the convention used in your course.) Some people would say that the limit is $\infty$, or sometimes $+\infty$. 
A: Hint:
$$\frac{n^3+2}{n^2+3}=\frac{n^3+3n-3n +2 }{n^2+3}= \frac{n(n^2+3)}{n^2+3} +\frac{2-3n}{n^2+3}=n +\frac{2-3n}{n^2+3} $$
A: 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 $+\infty$.  
