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Does $\lim_{x\rightarrow 0}\frac{c}{|x|}$ exist? I was under the impression that it does, because, even though $c/0$ is undefined, $c/|x|$ goes to infinity as $x$ approaches $0$.

However, my calcus textbook says that $\lim_{x\rightarrow a}\frac{f(x)}{g(x)}=\frac{\lim_{x\rightarrow a} f(x)}{\lim_{x\rightarrow a} g(x)} \operatorname{ if } g(x) \neq0$

Which seems to suggest that $\lim_{x\rightarrow a}\frac{c}{|x|}$ would likewise not exist if $a=0$.

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    @anon aw, now that you point that out, I see that I made an error transcribing the formula. It should read $\operatorname{...if} \lim_{x \rightarrow a} g(x) \neq0$. In any case I get what your saying. Thanks.2011-09-18

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The limit exists if and only if $c=0$, in which case the limit is $0$. You are correct that if $c\neq 0$, then $\lim_{x\to 0}\frac{c}{|x|} =\infty.$ However, when we write $\lim_{x\to a}f(x) = \infty$ in standard analysis, this says two things:

  • That the limit does not exist; and
  • That the reason why the limit does not exist is that the values of $f(x)$ are increasing without bound (that is, that for all $M\gt 0$ there exists $\delta\gt 0$ such that if $0\lt|x-a|\lt \delta$, then $f(x)\gt M$).

In other words, when the limit "is infinity", that is a special way for the limit to not exist. Similarly when we write that the limit "is" $-\infty$, we are really saying the limit does not exist, and explaining why it doesn't exist at the same time.


Added. I missed your final comment, and there @anon is quite right: the textbook is saying that if the limit of the numerator exists, and the limit of the denominator exists and is not zero, then the limit of the quotient exists and is the quotient of the limits.

This does not imply that if the conditions are not met then the limit does not exist. For example, with $\lim_{x\to 0}\frac{x}{x},$ the limit of the denominator is $0$, but the limit exists anyway (it's equal to $1$); however, the fact that limit is $1$ does not follow from trying to break this limit up into "limit of $x$ divided by limit of $x$" (that's not a valid step), but for other reasons (because $\frac{x}{x}$ and $1$ are equal everywhere except at $0$, and when we take the limit as $x\to 0$ what matters is what is happening near $0$ and not what happens at $0$).

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    No, though you make an interesting point. What I meant is that I was thinking that you could have $\lim_{x \rightarrow 2} h(x) = 0$ where $h(x)$ stops decreasing before $x = 2$, but then your comment made me realise that, if that were the case, then the limit could not be $0$.2011-09-18
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Well, this is a typical question I hear from my students. The case $c=0$ is trivial, isn't it? So, assume without loss of generality (why?) that $c=1$. Then $\lim_{x \to 0} \frac{1}{|x|}=+\infty,$ while $\lim_{x \to 0} \frac{-1}{|x|}=-\infty.$ However, what is meant here by "a limit exists"? Do you mean "exists in $\mathbb{R}$"? This is just a conventional issue.