does this function have a discontinuity at x = 0?\nf(x) = \\frac{|x|}{x}

Explanation:

Step1: Recall the definition of absolute - value

$|x|=\begin{cases}x, & x\geq0\-x, & x < 0\end{cases}$, so $f(x)=\frac{|x|}{x}=\begin{cases}\frac{x}{x}=1, & x>0\\frac{-x}{x}=-1, & x < 0\end{cases}$. The function is not defined at $x = 0$.

Step2: Check the limit as $x$ approaches 0

$\lim_{x\rightarrow0^{+}}f(x)=\lim_{x\rightarrow0^{+}}\frac{|x|}{x}=\lim_{x\rightarrow0^{+}}\frac{x}{x}=1$ and $\lim_{x\rightarrow0^{-}}f(x)=\lim_{x\rightarrow0^{-}}\frac{|x|}{x}=\lim_{x\rightarrow0^{-}}\frac{-x}{x}=-1$. Since $\lim_{x\rightarrow0^{+}}f(x)\neq\lim_{x\rightarrow0^{-}}f(x)$, the limit $\lim_{x\rightarrow0}f(x)$ does not exist.

Answer:

Yes