# $\lim_{x \to 0^-} \left(\frac{1}{x} - \frac{1}{|x|}\right),$ the limit does not exist.

Is this a good justification?

$$\lim_{x \to 0^-} \left(\frac{1}{x} - \frac{1}{|x|}\right),$$ the limit does not exist.

Because $$\lim_{x \to 0^-} \frac{1}{x} = -\infty$$ And $$\lim_{x \to 0^-} \frac{1}{|x|} = \infty.$$

We know that $$\lim_{x \to 0^-} \left(\frac{1}{x} - \frac{1}{|x|}\right) = -\infty.$$ Hence the limit does not exist.

• No. You know that $x<0$, so $\frac{1}{x}-\frac{1}{|x|} = \frac{2}{x}$, and $\lim_{x \uparrow 0} \frac{2}{x} = -\infty$. – copper.hat Sep 10 '13 at 6:17
• The idea is basically right. You probably saw that for $x$ negative and close to $0$, we get a very "big" negative number, since both parts are pulling in the same direction. However, your way of putting it makes it look as if you are doing algebraic manipulation on the symbol "$-\infty$." In general that's not a good thing to do, though it gives the right answer in this case. – André Nicolas Sep 10 '13 at 6:22

No, this is not a good justification. By similar logic, $4=\lim_{x\to\infty} 4= \lim_{x\to\infty} (x+4)-x=\lim_{x\to\infty} x+4 +\lim_{x\to\infty} -x=\infty-\infty$. It is only valid to break up a limit into a sum when the values are finite.

Hint: what is $|x|$ when $x<0$? Simplify the expression inside the parenthesis using this.

Your justification is not sufficient, since limit laws generally only apply when the limits are finite. The correct way to handle the limit is to note that since $x < 0$, $|x| = -x$. Hence,

$$\frac{1}{x} - \frac{1}{|x|} = \frac{1}{x} + \frac{1}{x} = \frac{2}{x} \to -\infty$$

as $x \to 0^-$.

You can only use that the limit of the difference is the difference of the limits when the limit is finite. To argue this note that we are approaching $0$ from the left. So all values of $x$ will be negative. Hence $|x|=-x$. So

$\lim_{x\to0^{-}}\bigg(\frac{1}{x}-\frac{1}{|x|}\bigg)=\lim_{x\to0^{-}}\frac{2}{x}=-\infty$

For a limit to exist the lateral limits have to be the same.

• By "lateral limit" do you mean the one-sided limit? Because this question is only about a one-sided limit. – user61527 Sep 10 '13 at 6:32
• no, 2-sided. I can see now. – Poli Tolstov Sep 10 '13 at 6:48