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

Can $$\lim_{x \to 0} \frac{x^3-7x}{x^3}$$ be rewritten as $$\lim_{x \to 0} \frac{x^3(1-7x^{-2})}{x^3}?$$

These two seem to give different answers. Please help me I'm really confused :S

share|cite|improve this question
up vote 1 down vote accepted

What you did is fine. The next step is to cancel the $x^3$ terms to get

$$\lim_{x \to 0} \frac{x^3 - 7x}{x^3} = \lim_{x \to 0} \frac{x^3 (1 - 7x^{-2})}{x^3} = \lim_{x \to 0} 1 - 7x^{-2}$$

Now, we have a polynomial over a polynomial and we get division by 0 but the numerator of the fraction is not 0. If it were a left or right hand limit, this would tell us the answer is $+\infty$ or $-\infty$ and we would only need to figure out which one it is. If it's a limit (not right or left hand), there is the possibility that the left and right hand limits differ (one is $+\infty$ and the other is $-\infty$) so the limit could not exist. But, as $x \to 0$, the whole thing $1 - 7x^{-2}$ is going to be negative no matter if you approach from the left or right. Therefore, the left and right hand limits are both $-\infty$, so that the limit itself is $-\infty$.

share|cite|improve this answer

They are indeed the same:

$$\lim_{x\to 0}\frac{x^3-7x}{x^3}=\lim_{x\to 0}\frac{x^3(1-7x^{-2})}{x^3}=\lim_{x\to 0}\left(1-\frac7{x^2}\right)\;,$$ which does not exist. However, it fails to exist in a relatively nice way: as $x\to 0$, the expression in parentheses gets increasingly negative without any bound, so we often write

$$\lim_{x\to 0}\left(1-\frac7{x^2}\right)=-\infty\;.$$

share|cite|improve this answer
Could you drop by the chat? – Pedro Tamaroff Sep 15 '12 at 1:48

Your rewrite is correct. The limit of the expression, in either form, does not exist. Alternately, if you allow $\infty$ or $-\infty$ as possible answers to a limit question, the limit is $-\infty$.

A direct way to calculate the limit is to note that if $x\ne 0$, then $$\frac{x^3-7x}{x^3}=1-\frac{7}{x^2}.$$

share|cite|improve this answer
Maybe it is worth to mention something related to "punctured neighborhood" in a more pedagogical manner, to explain why $$\mathop {\lim }\limits_{x \to 0} \frac{{{x^3} - 7x}}{{{x^3}}} = \mathop {\lim }\limits_{x \to 0} \left( {1 - \frac{7}{{{x^2}}}} \right)?$$ – Pedro Tamaroff Sep 15 '12 at 1:47
@PeterTamaroff: Thanks for the suggestion, I added something along the lines you mentioned, but without using the phrase "punctured neighborhood." – André Nicolas Sep 15 '12 at 1:53
Yes. Using the phrase "punctured neighborhood" wouldn't really help. – Pedro Tamaroff Sep 15 '12 at 1:56
@PeterTamaroff thanks for the term "punctured neighborhood", it gave me something to look up and "try" to understand. – yiyi Sep 15 '12 at 3:47

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.