Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
Nothing was wrong with multiplying top and bottom by $x^{-3}$, but $0^{-2} \neq 0$. – JavaMan Oct 20 '11 at 2:18
@DJC: It doesn't? Oh. Well, that explains a lot. Didn't know that thanks. Now that I think about I guess it is 1/0^2 and that doesn't equal zero thanks. – Dair Oct 20 '11 at 2:19
$0^{-2} = \frac{1}{0^2}$ which is undefined. – JavaMan Oct 20 '11 at 2:19
@DJC: If you put as an answer I would be happy to accept thank you. – Dair Oct 20 '11 at 2:20
For $x \ne 0$, $\frac{x^3-7x}{x^3}=1-\frac{7}{x^2}$. Now it should not be hard to see what happens as $x$ approaches $0$. – André Nicolas Oct 20 '11 at 3:11
up vote 4 down vote accepted

$0^{-2}$ 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.

share|cite|improve this answer
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. – The Chaz 2.0 Oct 20 '11 at 6:57

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.