<s>[INST] 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!
 [/INST] 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
$$\frac{n^3-3}{2n^2+n-1} \ge\frac{n^3-6n^2+9n}{4n^2-24n+36}=\frac{n(n-3)^2}{4(n-3)^2}=\frac n 4 \to \infty$$
</s>