Prove that $\displaystyle \lim_{x \to 0} \dfrac{|x|}{x}$ does not exist

Question

I thought absolute values were positive? Why is the there a negative $x$ in example $7$ in the attached picture. Can someone explain?

Answer 1

hint:Note that $x \to 0^{-}$ means that $x \to 0$, and $x < 0$, thus $|x| = -x$.

Answer 2

If you remember from the definition of the absolute value, $|x| = x$ if $x \geq 0$, and $|x| = -x$ if $x < 0$ (because if $x$ is less than zero, we take it to be negative, which makes it positive, for example, if $x = -3$, then $|x| = -(-3) = 3$, which is what we want).

So from the right side, we're taking $$\lim_{x \to 0^+} \frac{|x|}{x} = \lim_{x \to 0^+} \frac{x}{x} = \lim_{x \to 0^+} 1 = 1$$ and from the right side, we're taking $$\lim_{x \to 0^-} \frac{|x|}{x} = \lim_{x \to 0^-} -\frac{x}{x} = \lim_{x \to 0^-} -1 = -1$$ And remember that we can only say the limit exists if it approaches the same value from both sides.

Answer 3

It might help to think of the absolute value of a number as the distance of a number from zero on the number line. For instance, the absolute value of $7$ and $-7$ is $7$. Why? because the distance of $7$ and $-7$ from $0$ are both $7$.

To answer your question as to why $x$ is negative as it approaches $0$ from the left side, know that for any $x\in\mathbb{R}$, the absolute value of $x$ denoted $|x|$ is defined by

$$|x| = \begin{cases}-x &\text{if $x<0,$} \\ 0 &\text{if $x=0,$} \\x &\text{if $x>0.$} \\\end{cases}$$

Since $x$ is approaching from the left side, we must have that $x<0$. By definition of absolute value, we use $-x$.

Your confusion with the absolute value being positive all the time lies in thinking about the distance of numbers $x$ from $0$. For $x<0$, the $\textbf{distance}$ of $x$ from $0$ must of course be positive, but by definition, the $\textbf{sign}$of $x$ is negative for $x<0$.

Answer 4

The absolute value is never negative -- you're confusing the minus sign with it being negative. Since $x<0$, we have that $|x| = -x > 0$. Hence we get when approaching from the left side $|x|/x = -x/x = -1$.

For example when $x = -0.01$, then $|x| = -(-0.01) = 0.01$ and $|x|/x = 0.01/(-0.01) = -1.$

Answer 5

Consider $\varepsilon=1/2$; if the limit exists and equals $l$, you could find $\delta>0$ such that, for $0<|x|<\delta$, $$ \left|\frac{|x|}{x}-l\right|<\frac{1}{2} $$ However, for $x=\delta/2$, we have $$ \frac{|\delta/2|}{\delta/2}=1 $$ so we must have $|1-l|<1/2$, that is $-1/2<l-1<1/2$ or $$ \frac{1}{2}<l<\frac{3}{2}. $$ But, for $x=-\delta/2$, we have $$ \frac{|-\delta/2|}{-\delta/2}=-1 $$ so we must have $|-1-l|<1/2$, that is $-1/2<l+1<1/2$ or $$ -\frac{3}{2}<l<-\frac{1}{2}. $$ Since we can't have both $l>1/2$ and $l<-1/2$, we conclude that $l$ doesn't exist.