MathAdmin at 15:57, 4 December 2017

2017-12-04T15:57:13Z

MathAdmin: Created page with " Discuss, without graphing, if the following function is continuous at $x=0.$ :: $f(x) = \left\{ \beg..."$

2017-04-10T16:29:40Z

Created page with " Discuss, without graphing, if the following function is continuous at <math style="vertical-align: 0px">x=0.</math> ::<math>f(x) = \left\{ \beg..."

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Discuss, without graphing, if the following function is continuous at  <math style="vertical-align: 0px">x=0.</math>

::<math>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.
</math>

If you think  <math style="vertical-align: -4px">f</math>  is not continuous at  <math style="vertical-align: -4px">x=0,</math>  what kind of discontinuity is it?

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|<math style="vertical-align: -5px">f(x)</math>  is continuous at  <math style="vertical-align: 0px">x=a</math>  if
|-
|       <math style="vertical-align: -14px">\lim_{x\rightarrow a^+}f(x)=\lim_{x\rightarrow a^-}f(x)=f(a).</math>
|}

'''Solution:'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We first calculate  <math style="vertical-align: -14px">\lim_{x\rightarrow 0^+}f(x).</math>  We have
|-
|
       <math>\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}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we calculate  <math style="vertical-align: -14px">\lim_{x\rightarrow 0^-}f(x).</math>  We have
|-
|
       <math>\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}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|Since
|-
|
        <math style="vertical-align: -15px">\lim_{x\rightarrow 3^+}f(x)=\lim_{x\rightarrow 3^-}f(x)=-1,</math>
|-
|we have
|-
|        <math>\lim_{x\rightarrow 3} f(x)=-1.</math>
|-
|But,
|-
|        <math>f(0)=0\ne \lim_{x\rightarrow 3} f(x).</math>
|-
|Thus, <math style="vertical-align: -5px">f(x)</math>  is not continuous.
|-
|It is a jump discontinuity.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|       <math style="vertical-align: -5px">f(x)</math>  is not continuous. It is a jump discontinuity.
|}
[[009A_Sample_Final_3|'''Return to Sample Exam''']]

@@ Line 61: / Line 61: @@
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -15px">\lim_{x\rightarrow 3^+}f(x)=\lim_{x\rightarrow 3^-}f(x)=-1,</math>
+&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -15px">\lim_{x\rightarrow 0^+}f(x)=\lim_{x\rightarrow 0^-}f(x)=-1,</math>
 |-
 |we have
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{x\rightarrow 3} f(x)=-1.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{x\rightarrow 0} f(x)=-1.</math>
 |-
 |But,
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>f(0)=0\ne \lim_{x\rightarrow 3} f(x).</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>f(0)=0\ne \lim_{x\rightarrow 0} f(x).</math>
 |-
 |Thus, <math style="vertical-align: -5px">f(x)</math>&nbsp; is not continuous.

009A Sample Final 3, Problem 4 - Revision history

MathAdmin at 15:57, 4 December 2017

MathAdmin: Created page with " Discuss, without graphing, if the following function is continuous at x=0. ::f(x) = \left\{ \beg..."

MathAdmin: Created page with " Discuss, without graphing, if the following function is continuous at $x=0.$ :: $f(x) = \left\{ \beg..."$