Take the 2-minute tour ×

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|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. –  user667648 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. –  user667648 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
add comment

2 Answers 2

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|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
add comment

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|improve this answer
add comment

Your Answer


