Can I compute $\lim_{n \to \infty}(\frac{n-1}{3n})^n$ this way? Can I do the following to compute the limit: $\lim_{n \to \infty}(\frac{n-1}{3n})^n = \lim_{n \to \infty}(\frac{n}{3n}-\frac{1}{3n})^n = \lim_{n \to \infty}(\frac{1}{3}-\frac{1}{3n})^n = (\frac{1}{3}-0)^\infty = (\dfrac{1}{3})^\infty = 0$

No, you can't do that. You make the mistake here:
$$
\lim_{n \to \infty}(\frac{1}{3}-\frac{1}{3n})^n {\color{red}=} (\frac{1}{3}-0)^\infty
$$
What you can do is to let $y = \left(\frac{n-1}{3n}\right)^n$ and then consider
$$
\ln(y) = n\ln\left(\frac{n-1}{3n}\right) = \frac{\ln((n-1)/3n)}{n^{-1}}.
$$
Then you can use L'Hopital's rule to find the limit of $\ln(y)$ as $n$ approaches infinity. Then
$$
\lim_{n\to \infty} y = \lim_{n \to \infty} e^{\ln(y)}.
$$
