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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 $\frac{x^{-3}}{x^{-3}}$ ?

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$0^{-2}$ 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.

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    Notice that $\lim_{x \to 0} \frac{1}{x} $ does not exist, since some of these "infintesimally close to 0" numbers are negative, and some are positive.2011-10-20
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Notice how when you multiplied by $\frac{x^{-3}}{x^{-3}}$ and got $1-7x^{-2} = 1-\frac{7}{x^2}$, the limit $\lim_{x\rightarrow 0} -\frac{7}{x^2} = -\infty$ then $\lim_{x\rightarrow 0} 1 - \frac{7}{x^2} = -\infty$.

Remember that obtaining limits DOES NOT CONSIST OF REPLACING x BY SOME VALUE. There usually is some thinking involved.