Limit of $f(x)=x/|x|$ as $x\to 0$

Question

Consider the function $f$ defined in $\mathbb{R}\setminus\{0\}$ by $f(x)=\frac{x}{|x|}$. Show that $\displaystyle\lim_{x\to 0}f(x)$ doesn't exist.

My attempt:

Note that $f(x)=1$ for $x>0$ and $f(x)=-1$ for $x<0$.

Suppose that $\displaystyle\lim_{x\to 0}f(x)=L$ exists.

Applying the definition with $\varepsilon =1$, there should be $\delta>0$ such that $$0<|x-0|<\delta\implies |f(x)-L|<1.$$

Taking $x=\delta/2$, implies $|1-L|<1$, then $-1<-1-L<1$, which implies $1-L<1\iff 0<L$

Now, taking $x=-\delta/2$ implies $|-1-L|<1,$ then $-1<-1-L<1,$ which implies $-1<-1-L\iff L<0$. (Contradiction)

Is this correct?

Answer 1

Your first observation for right and left limit

$$\lim_{x\to 0^+} \frac{x}{|x|}=1 \neq \lim_{x\to 0^-} \frac{x}{|x|}=-1 $$

suffices to prove that the limit at $x=0$ doesn't exist by uniqueness theorem for limits.

Answer 2

Actually this is signum function. $$f(x)=\frac{|x|}{x}$$ is its identical twin.

Answer 3

You can also get the contradiction in one fell swoop from the strict inequality step in

$$\begin{align} 2&=|1-(-1)|\\ &=|f(\delta/2)-f(-\delta/2)|\\ &=|(f(\delta/2)-L)+(L-f(-\delta/2))|\\ &\le|f(\delta/2)-L|+|L-f(-\delta/2)|\\ &\lt1+1\\ &=2 \end{align}$$

where the key step invokes the triangle inequality $|a+b|\le|a|+|b|$.