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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

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3 Answers 3

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 $-\infty$.

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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 $$\frac{x^3-7x}{x^3}=1-\frac{7}{x^2}.$$

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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

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

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

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Could you drop by the chat? –  Pedro Tamaroff Sep 15 '12 at 1:48

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