The limit of $\frac{n^3-3}{2n^2+n-1}$ I have to find the limit of the sequence above. 
Firstly, I tried to multiply out $n^3$, as it has the largest exponent.
$$\lim_{n\to\infty}\frac{n^3-3}{2n^2+n-1} = 
\lim_{n\to\infty}\frac{n^3(1-\frac{3}{n^3})}{n^3(\frac{2}{n} + \frac{1}{n^2} - \frac{1}{n^3})} = 
\lim_{n\to\infty}\frac{1-\frac{3}{n^3}}{\frac{2}{n} + \frac{1}{n^2} - \frac{1}{n^3}}$$
$$
\begin{align}
\lim_{n\to\infty}1-\frac{3}{n^3} = 1 \\[1ex]
\lim_{n\to\infty}\frac{2}{n} + \frac{1}{n^2} - \frac{1}{n^3} = 0 \\[1ex]
\lim_{n\to\infty}\frac{n^3-3}{2n^2+n-1} = \frac{1}{0}
\end{align}
$$
Then, after realizing $\frac{1}{0}$ might not be a plausible limit, I tried to multiply out the variable with the largest exponent in both the dividend and the divisor.
$$\lim_{n\to\infty}\frac{n^3-3}{2n^2+n-1} = \lim_{n\to\infty}\frac{n^3(1 - \frac{3}{n^3})}{n^2(2 + \frac{1}{n} - \frac{1}{n^2})} = 
\lim_{n\to\infty}n\cdot\frac{1 - \frac{3}{n^3}}{2 + \frac{1}{n} - \frac{1}{n^2}}$$
$$
\begin{align}
\lim_{n\to\infty}1-\frac{3}{n^3} = 1 \\
\lim_{n\to\infty}2 + \frac{1}{n} - \frac{1}{n^2} = 2 \\
\lim_{n\to\infty}\frac{n^3-3}{2n^2+n-1} = \frac{1}{2} \\
\lim_{n\to\infty}n = \infty
\end{align}
$$
So, my questions about this problem:


*

*Could $\frac{1}{0}$ be a valid limit? 

*Does $\infty\cdot\frac{1}{2}$
equal to $\infty$?  

*In conclusion, what is the limit of the sequence
above? $\infty?$


Thank you!


Could $\frac{1}{0}$ be a valid limit?
You should be asking instead :"Is the following true?"  $$\lim_{n \rightarrow \infty} \frac{f(n)}{g(n)} = 0$$
where $\lim_{n \rightarrow \infty} f(n) = K $, where $K$ is finite and $\lim_{n \rightarrow \infty} g(n) = +\infty$. The answer is yes. This is one of the limit rules.



Does $\infty \frac{1}{2}$ equal to $\infty$?
You should be asking instead:"Is the following true?"  $$\lim_{n \rightarrow \infty} K f(n)= \infty$$
where $K$ is finite and $\lim_{n \rightarrow \infty} f(n) = +\infty$. The answer is yes. This is one of the limit rules.



In conclusion, what is the limit of the sequence above?
Following the same steps as you did, in both approaches, I have nothing to add. The limit is $\infty$.

