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
$\lim _{x\rightarrow 3^{+}}f(x)=\lim _{x\rightarrow 3^{-}}f(x)=-1,$
we have
$\lim _{x\rightarrow 3}f(x)=-1.$
But,
$f(0)=0\neq \lim _{x\rightarrow 3}f(x).$
Thus, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)}   is not continuous.
It is a jump discontinuity.

Final Answer:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)}   is not continuous. It is a jump discontinuity.

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