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

Basically, no.
When you go from the third step to the fourth one, you make a false assumption. The same one that confuses people with $e$: $$\lim_{n\to\infty}\left(1+\frac1n\right)^n=e\neq \lim_{n\to\infty}\left(1+\frac1\infty\right)^\infty=1
$$
In simple terms, the reason why this is so is that if you expand the expression of the form $(a+b/n)^n$ you will get lot of stuff. And there's some stuff that quickly goes to $0$, but another part will converge to a real value since there will be some cancellations (when you set $b/n=0$ you basically remove all those cancellations and so you're left with only $a^n$). Example: $\lim\limits_{n\to\infty}(a+b/n)^n$ will at least be $2$ in the case of $a=b=1$ since by expanding $(1+1/n)^n$ you will get: $1^n+n\cdot1/n+\text{lot of stuff}=2+\text{lot of stuff}$ by the binomial theorem.
I hope this helps.
Best wishes, $\mathcal H$akim.
