The limit of $\frac{n^3-3}{2n^2+n-1}$

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 $\infty$.

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