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

One thing to remember when you're taking limits involving infinity is that you're walking a very fine line and it's easy to make assumptions and forget that $\infty$ is a completely different animal than any other number. 
$$\lim_{n \to \infty} \left( \frac{n-1}{3n}\right)^n$$
What does this look like to you? Doesn't that form make you a little bit queasy? 
Look for some patterns:
$$\lim_{n \to \infty} \left( \underbrace{\frac{\textbf{1}n-1}{3n}}\right)^\underbrace{n}$$
Hmm, maybe before we go trodding off, we can do some algebra. 
$$\lim_{n \to \infty} 3^{-n} \left( \frac{n-1}{n}\right)^n = \lim_{n \to \infty} \frac{1}{3^n} \left( \frac{n-1}{n}\right)^n$$
Think

Now, I don't want to cursorily glaze over a component of this. I think it would help your understanding to do an altered form of the problem, which in the end may help your understanding. Let's suppose that 3n term at the bottom didn't exist. Just for a second, let's focus on:
$$\lim_{n\to\infty}\left( \frac{n-1}{n}\right)^n$$
Notice that as $n$ becomes arbitrarily large it smashes that puny little $-1$ term in the numerator, it's like a pebble of sand in comparison to the size of the universe. Imagine the largest number that you can think of for $n$, and visualize it becoming larger and larger.
Well, as one massively large thing goes to divide another massively large thing, the leading coefficient is what matters, and $\frac{1}{1} = 1$ This whole thing looks like it's going to 1. So:
This leads us to a type of indeterminate form $1^\infty$, so what do we do? Recall that we need to transform this type of indeterminate form using e and the natural logarithm:
$$\huge{ e^{ \lim_{n\to\infty} n \cdot \ln\left( \frac{n-1}{n} \right)} }$$
Now we have another indeterminate form of type $0\cdot\infty$. This is because as $n$ goes to $\infty$, the natural logarithm of $\frac{n-1}{n}$ goes to 0. We have to apply another transformation: 
We want to get this into a quotient form. Recall that for some $f(x)g(x)$ we can transform it into a product by doing $\large{\frac{f(x)}{\frac{1}{g(x)}}}$. What we're going to do next is take that pesky $n$ as it's approaching infinity and invert it to become our $\frac{1}{g(x)}$ and take our $f(x)$ to be the rest. Let $t$ be $\frac{1}{n}$:
$$ \lim_{n\to\infty} n \cdot \ln\left( \frac{n-1}{n} \right) =  \lim_{t\to0} \frac{\ln\left( \frac{1}{t} - 1 \right)t}{t}$$
So: 
$$\huge{ e^{\lim_{t\to0} \frac{\ln\left( \frac{1}{t} - 1 \right)t}{t}} } $$
Which is of indeterminate form $\frac{0}{0}$, so we apply L'Hopital's rule:
$$\large{ e^{\lim_{t\to0} \frac{1}{t-1} } = \frac{1}{e} } $$
Going Back

So now that we know that: 
$$\lim_{n \to \infty} \left( \frac{n-1}{3n}\right)^n = \frac{1}{e}$$
So, if we go back to the step where I did some algebraic manipulation of the problem: 
$$\lim_{n \to \infty} \frac{1}{3^n} \left( \frac{n-1}{n}\right)^n$$
Can you feel how close we are? It becomes to readily obvious that: 
$$ \huge{ \lim_{n \to \infty}  \frac{{\underbrace{ \left( \frac{n-1}{n}\right)^n }_{\frac{1}{e}}}}{3^n} } = 0$$
And as $n$ goes to $\infty$, it smashes that constant in the numerator...
Does this make more sense to you now?
