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.

The book says that $$\lim_{x \rightarrow 0^{-}} \left( \frac{1}{x} - \frac{1}{|x|} \right) \mbox{does not exist}$$
But, given any $M \lt 0$ of large magnitude, if I choose $\delta = \frac{-x^{2}M}{2}$ then any value of x where $|x-0|< \delta$ and $x <0$ (as we are coming from the left) will lead to $\left( \frac{1}{x} - \frac{1}{|x|} \right) < M$. To me, that says that my text book is incorrect in saying that this limit "d.n.e."

I'm a little bothered that my $\delta$ depends on $x$, but I tried a few numerical examples and it worked fine. Perhaps the function is not uniformly continuous when $x \lt 0$? I have not done enough work to answer that question yet.

Maybe the book meant to say
$$\lim_{x \rightarrow 0} \left( \frac{1}{x} - \frac{1}{|x|} \right) \mbox{does not exist?}$$ Or maybe I have missed something elementary.

share|improve this question
Since you're taking the limit as $x$ approaches $0$ from the left, doesn't $\frac{1}{x} - \frac{1}{|x|}$ simplify to $\frac{2}{x}$, whose limit aproaching $0$ from the left doesn't exist? –  Jason DeVito Feb 11 '11 at 4:27
$M$ is a very large negative number. If $\big(\frac{1}{x}-\frac{1}{|x|}\big) < M$, then it is unbounded as $x \rightarrow 0^-$. –  Brandon Carter Feb 11 '11 at 4:31
It's un-bounded and appears to be negative infinity, based on my proof, which would mean that the limit exists. –  futurebird Feb 11 '11 at 4:36
Jason your way of simplifying is much more simple than mine. Duh. –  futurebird Feb 11 '11 at 4:38
@a little don: $\delta$ should not depend on $x$; it seems you might be confusing a bit with uniform continuity. In general, to show a function is continuous at $a$ your $\delta$ may depend on $a$ and $\epsilon$ (but not on $x$); the function is uniformly continuous if $\delta$ does not depend on $a$. –  Arturo Magidin Feb 11 '11 at 20:12

1 Answer 1

up vote 4 down vote accepted

One problem with your $\delta$ is that $|x|\lt \frac{-x^2M}{2}$ with $x\neq0$ implies $|x|\gt-\frac{2}{M}$. Having this positive lower bound on $|x|$ means that you are not actually approaching $0$. As Jason DeVito indicates, taking advantage of the fact that $x\lt0$ to write $|x|=-x$ makes it easier to see why the limit doesn't exist.

In a comment you just mentioned that the limit is $-\infty$. That is true, but then there is a divide in terminology as to what having a limit means. Infinite limits can be dealt with in terms of convergence in the extended reals, $[-\infty,+\infty]$, but often $\pm\infty$ are treated separately with respect to limits, and the existence of a limit (without further qualification) often only refers to existence of a limit in the real number system, $(-\infty,+\infty)$.

share|improve this answer
Thanks for finding the problem in my proof. The book talks about infinite limits "existing" it says, for example, that lim (x-->0) 1/x = dne, while lim (x-->0) 1/(x^2) = infinity. –  futurebird Feb 11 '11 at 4:45
@a little don: Hmm, that is confusing. There may have been a minor oversight in editing at one or another part of the book. –  Jonas Meyer Feb 11 '11 at 4:48
@a little don: in most of the calculus texts that I've seen, they will write that a limit "equals infinity" (or negative infinity) as a shorthand for saying that the limit does not exist, but does not exist in a specific way that is further described as the expression increasing (or decreasing) without bound. –  Isaac Feb 11 '11 at 14:00

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.