Classifying the discontinuity of |x|/x

Text

Classify the points of discontinuity of $f(x) = \dfrac{|x|}{x}$.

Solution

The function is defined for $x\ne 0$. Making the absolute value explicit: $f(x) = \frac{|x|}{x} = \begin{cases} \dfrac{x}{x}=1 & x>0 \\[4pt] \dfrac{-x}{x}=-1 & x<0 \end{cases}$ So $f$ is constantly $+1$ to the right of zero and $-1$ to the left. The one-sided limits at $x=0$ are $\lim_{x\to 0^+}f(x)=1, \qquad \lim_{x\to 0^-}f(x)=-1.$ Both are finite but different: at $x=0$ there is a discontinuity of $\boxed{\text{first kind (jump)}}$, of amplitude $|1-(-1)|=2$. It is the only point of discontinuity, since elsewhere $f$ is constant and therefore continuous.

Links

Topics: Continuity
Concepts: Discontinuity · First-kind discontinuity · Right and left limits · Absolute value
Skills: Calculating limits · Reasoning by cases
Exercise types: Reading a graph