Q:

What is wrong with my logic?

So I was trying to do some problems from this website. And on Problem number 10 I tried to do the following:
$$\lim_{x \to 0} \frac{x^3-7x}{x^3}$$
Multiply everything by $\frac{x^{-3}}{x^{-3}}$
$$\lim_{x \to 0} \frac{x^3-7x}{x^3}\times\frac{x^{-3}}{x^{-3}}$$
Which I got equals:
$$\lim_{x \to 0} \frac{1-7x^{-2}}{1}$$
Plug in $0$ for $x$ and I get:
$$\frac{1}{1} = 1$$
But, the answer according to the website is $-\infty$. (And therefore no limit exists). What was wrong about multiply by $\frac{x^{-3}}{x^{-3}}$ ?

A:

$0^{-2}$ is $\frac1{0^2}$ which is $\frac1{0}$. Now, normally this would be a divide by zero right? Well, with limits it's not technically $0$, it's actually a very tiny number that's infintesimally close to $0$. So when you divide $1$ by some itty bitty number, you get a very massive number. As you bring that number that's very close to $0$ ever closer, the result grows ever larger. It grows infinitely large and thus to infinity. The negative arises from the negative cooeficient if I recall correctly.

