Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $a=0$.

share|improve this question
2  
In standard analysis, infinity isn't a real number, and we only say a limit exists if it is a real number. –  anon Sep 18 '11 at 4:51
1  
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). –  anon Sep 18 '11 at 5:03
1  
@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.) –  anon Sep 18 '11 at 5:16
1  
@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, and it is invalid reasoning. –  Arturo Magidin Sep 18 '11 at 5:19
2  
@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.) –  anon Sep 18 '11 at 5:38

2 Answers 2

up vote 5 down vote accepted

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 $0$).

share|improve this answer
    
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? –  Matt Munson Sep 18 '11 at 5:06
    
@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 $(\lim f(x))/(\lim g(x)) = 1/2$. –  Arturo Magidin Sep 18 '11 at 5:16
    
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? –  Matt Munson Sep 18 '11 at 5:22
    
@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. –  Arturo Magidin Sep 18 '11 at 5:33
    
@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 $2$. –  Arturo Magidin Sep 18 '11 at 5:35

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.

share|improve this answer

Your Answer

 
discard

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.