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!
 A: 
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 $\infty$.

A: The second way is preferable and since the product "$\infty \cdot \frac12$" is not an indeterminate form from here
$$\lim_{n\to\infty}n\cdot\frac{1 - \frac{3}{n^3}}{2 + \frac{1}{n} - \frac{1}{n^2}}$$
we can conclude that the sequence diverges.
Note that also with the first method from here
$$\lim_{n\to\infty}\frac{1-\frac{3}{n^3}}{\frac{2}{n} + \frac{1}{n^2} - \frac{1}{n^3}}$$
since also  "$\frac1 0$" is not an indeterminate form and the denominator is $>0$ we can conclude that the sequence diverges at $\infty$.
In both cases, what we cannot do is to calculate separetely the limit of each part and then make a direct calculation (that's allowed only is the final operation in well defined).
As an alternative since eventually


*

*$n^3-3\ge n^3-6n^2+9n \iff6n^2-9n-3\ge 0$

*$2n^2+n-1 \le 4n^2-24n+36\iff2n^2-25n+35 \ge 0$


by squeeze theorem we have
$$\frac{n^3-3}{2n^2+n-1} \ge\frac{n^3-6n^2+9n}{4n^2-24n+36}=\frac{n(n-3)^2}{4(n-3)^2}=\frac n 4 \to \infty$$
A: I admire your efforts, another approach is
$$\lim_{n\to\infty}\frac{n^3-3}{2n^2+n-1} = 
\lim_{n\to\infty}\left(\frac12n-\frac14+ \dfrac{\frac34n - \frac{13}{4}}{2n^2+n-1} \right) =\infty$$
A: 
So, my questions about this problem:

*

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


No, since $\frac10$ is not defined.


*

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


There are the "extended real numbers" $\mathbb{R}\cup\{\pm\infty\}$
Where you can define this.
Note, that $\infty$ is in general not a number. So you can not do some calculations with it.
You can check out: https://en.wikipedia.org/wiki/Extended_real_number_line
For more informations.


*

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

Yes, the limit is $\infty$. Or in other words the sequence does not converge.
A: *

*No, it's not valid, it is not defined. Consider two sequences: $(a_n)_{n\in\mathbb{N}} = \frac{1}{2^{-n}}$ and $(b_n)_{n\in\mathbb{N}} = \frac{1}{-2^{-n}}$. You could say that limit of both of them is $\frac{1}{0}$, since $\pm2^{-n} \to 0$, but does this make sense? Do these sequences have the same limit?

*Saying that $\infty\cdot\frac{1}{2} = \infty$ is not very accurate, but your intuition is good. In general, for any sequence $(a_n)_{n\in\mathbb{N}}$ divergent to $+\infty$ and for any $x>0$, you have $\lim_{n\to\infty} x\cdot a_n = +\infty$. What happens for $x < 0$?

*It is $+\infty$.

A: All you had expected, was solved by other solutions. I'd like to note some points. Maybe it has another taste. We know that for every natural number $n ≥ 1$, if $P(x)$ is a polynomial of degree $n$ with leading coefficient $a$, then $$\lim_{x\to\infty}P(x)=\infty (-\infty)$$ whether $a$ is positive (negative). This means equally that, the limit of such polynomial at infinity, can be evaluated by the term which identify the order of $P(x)$. So at infinity, $$n^3-3\approx n^3,~~~2n^2+n−1\approx 2n^2$$ Now do the limit with the two result. 
A: You have done everything.I just want to represent it graphically.Then you will answer your own question.

Now,you tell me.what are you seeing here?
And ofcourse the answer of your your questions is simply "yes".
