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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.

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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. –  a little don Feb 11 '11 at 4:36
    
Jason your way of simplifying is much more simple than mine. Duh. –  a little don 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, $(-\infty,+\infty)$.

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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. –  a little don 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
3  
@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

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