# 009A Sample Final 3, Problem 4

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Discuss, without graphing, if the following function is continuous at  ${\displaystyle x=0.}$

${\displaystyle f(x)=\left\{{\begin{array}{lr}{\frac {x}{|x|}}&{\text{if }}x<0\\0&{\text{if }}x=0\\x-\cos x&{\text{if }}x>0\end{array}}\right.}$

If you think  ${\displaystyle f}$  is not continuous at  ${\displaystyle x=0,}$  what kind of discontinuity is it?

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

Solution:

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

${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow 0^{+}}f(x)}&=&\displaystyle {\lim _{x\rightarrow 0^{+}}x-\cos x}\\&&\\&=&\displaystyle {0-\cos(0)}\\&&\\&=&\displaystyle {-1.}\end{array}}}$

Step 2:
Now, we calculate  ${\displaystyle \lim _{x\rightarrow 0^{-}}f(x).}$  We have

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

Step 3:
Since

${\displaystyle \lim _{x\rightarrow 3^{+}}f(x)=\lim _{x\rightarrow 3^{-}}f(x)=-1,}$

we have
${\displaystyle \lim _{x\rightarrow 3}f(x)=-1.}$
But,
${\displaystyle f(0)=0\neq \lim _{x\rightarrow 3}f(x).}$
Thus, ${\displaystyle f(x)}$  is not continuous.
It is a jump discontinuity.

${\displaystyle f(x)}$  is not continuous. It is a jump discontinuity.