# 009A Sample Final 3, Problem 4

Discuss, without graphing, if the following function is continuous at  $x=0.$ $f(x)=\left\{{\begin{array}{lr}{\frac {x}{|x|}}&{\text{if }}x<0\\0&{\text{if }}x=0\\x-\cos x&{\text{if }}x>0\end{array}}\right.$ If you think  $f$ is not continuous at  $x=0,$ what kind of discontinuity is it?

Foundations:
$f(x)$ is continuous at  $x=a$ if
$\lim _{x\rightarrow a^{+}}f(x)=\lim _{x\rightarrow a^{-}}f(x)=f(a).$ Solution:

Step 1:
We first calculate  $\lim _{x\rightarrow 0^{+}}f(x).$ We have

${\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow 0^{+}}f(x)}&=&\displaystyle {\lim _{x\rightarrow 0^{+}}x-\cos x}\\&&\\&=&\displaystyle {0-\cos(0)}\\&&\\&=&\displaystyle {-1.}\end{array}}$ Step 2:
Now, we calculate  $\lim _{x\rightarrow 0^{-}}f(x).$ We have

${\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow 0^{-}}f(x)}&=&\displaystyle {\lim _{x\rightarrow 0^{-}}{\frac {x}{|x|}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow 0^{-}}{\frac {x}{-x}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow 0^{-}}-1}\\&&\\&=&\displaystyle {-1.}\end{array}}$ Step 3:
Since

$\lim _{x\rightarrow 0^{+}}f(x)=\lim _{x\rightarrow 0^{-}}f(x)=-1,$ we have
$\lim _{x\rightarrow 0}f(x)=-1.$ But,
$f(0)=0\neq \lim _{x\rightarrow 0}f(x).$ Thus, $f(x)$ is not continuous.
It is a jump discontinuity.

$f(x)$ is not continuous. It is a jump discontinuity.