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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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    In standard analysis, infinity isn't a real number, and we only say a limit exists if it is a real number.2011-09-18
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    Your calculus textbook does warn you about the case $g(x)\neq 0$ so why consider it?2011-09-18
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    @anon Alright. Is that were the distinction between a limit and an 'infinite limit' comes in?2011-09-18
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    Matt: Yes, see @Arturo's answer. Also do note that your idea of what the textbook 'suggests' is logically fallacious: it makes the assumption an implication is equivalent to its converse. In other words, just because $L\Leftarrow G$ (as the textbook states) doesn't mean $\mathrm{not}(G)\implies\mathrm{not}(L)$ (as you inferred).2011-09-18
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    @anon I'm not familiar with your notation.2011-09-18
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    @Matt: Sorry, I shouldn't have been presumptuous. Say $L$ is the limit formula and $G$ is the hypothesis $g(x)\ne0$. Then you appear to be saying "$L$ if $G$" (the textbook statement) suggests that "the formula $L$ is false when the hypothesis $G$ is false" (which is the converse of the textbook statement). In general, "IF" statements aren't logically equivalent to their converses. (I'm just providing this as tangential commentary about logic.)2011-09-18
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    @Matt: $L\Leftarrow G$ means "$L$ holds when $G$ holds"; $\mathrm{not}(G)\Rightarrow \mathrm{not}(L)$ means "if $G$ does not hold, then $L$ does not hold". In other words, you are being told that **if** A, **then** B; you are trying to deduce from this that if $A$ is false, then $B$ will be false. This is a classic fallacy called [denying the antecedent](http://en.wikipedia.org/wiki/Denying_the_antecedent), and it is invalid reasoning.2011-09-18
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    @anon Ah, ok. So you're saying that if $g(x)\neq0$ then $L$ is definitely true, but if $g(x) = 0$ we cannot be sure that $L$ is true, though it still might be.2011-09-18
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    @Matt: Close. I interpreted "$g(x)\ne0$" to mean "$g$ is never $0$," so what I'm saying is just because $L$ is true when $g$ is never $0$, doesn't mean you can conclude its false if $g$ *can* be zero. (Note that saying $g$ can equal zero for some $x$ inputs isn't the same as saying $g(x)=0$, which means it's identically zero.)2011-09-18
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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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    What if, for example, $g(x) = {1 \operatorname{if} x \neq 0, 0 \operatorname{if} x = 0}$ and $g(x) = {2 \operatorname{if} x \neq 0, 0 \operatorname{if} x = 0}$. Would $\lim_{x\rightarrow 0}\frac{f(x)}{g(x)}$ then exist?2011-09-18
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    @Matt: You defined two $g$s. I think the first is meant to be $f$, the second $g$; if $f(x)=1$ for all $x\neq 0$, then regardless of what happens to $f$ at $0$ we have $\lim_{x\to 0}f(x) = 1$; and if $g(x)=2$ for all $x\neq 0$, then regardless of what happens with $g$ at $0$, $\lim_{x\to 0}g(x)=2$. So in this case, the limit law you quote applies and the limit of $f(x)/g(x)$ as $x\to 0$ would be $(\lim f(x))/(\lim g(x)) = 1/2$.2011-09-18
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    There is an exercise in the text where it asks whether $\lim_{x \rightarrow 2}\frac{g(x)}{h(x)}$ exists, where $\lim_{x \rightarrow 2}g(x)=-2$ and $\lim_{x \rightarrow 2}h(x)=0$. And the answer key says that this limit doesn't exist. However, it seems like, by what you are saying, that can't be concluded. Would that be right?2011-09-18
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    @Matt: There is a difference between saying "you cannot conclude this from *this* limit law" and "it cannot be concluded." The logic you attempted to use in your original post is invalid and so you cannot use it to reach that conclusion. However, one can argue **differently** to show that the limit does not exist. Specifically: suppose the limit of $g/h$ existed. Then, since the limit of $g$ exists and is nonzero, then you would also know that the limit of $(g/h)/g = (1/h)$ exists. But the limit of $1/h$ cannot exist, because $h\to 0$: the quotient grows in absolute value without bound.2011-09-18
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    @Matt: (cont) That contradiction arises from assuming that the limit of $g/h$ *does* exist, so from **that** contradiction we can deduce that the limit does not exist. Alternatively: the limit of $|g/h|$ grows without bound, because as $x\to 2$, $|g(x)|$ is approaching $2$, while $|h(x)|$ approaches $0$ and is positive, so the quotient is getting larger and larger and larger as $x$ gets closer to $2$.2011-09-18
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    Ok, that ties it all together now. So then if $\lim_{x \rightarrow 2} h(x) = 0$, then $h$ cannot settle to any value $v$ as it moves infinitesimally closer to 2, because, if it did, then the limit by definition would be $v$, and thus $g/h$ is unbounded as x approaches $0$. Thanks for the explanation.2011-09-18
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    @Matt: Was "then $h$ cannot settle to any value" meant to be "then $g/h$ cannot settle to any value"? If so, that's true when the limit of $g$ exists and is nonzero. When the limit of $g$ exists but is equal to $0$, all best are off: what happens to $g/h$ cannot be determined merely from the information that both $g$ and $h$ have limit $0$ as $x\to 2$.2011-09-18
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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.