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| | <span class="exam">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? | | <span class="exam">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? |
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| | + | [[009A Sample Final 3, Problem 4 Solution|'''<u>Solution</u>''']] |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Foundations:
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| − | |-
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| − | |<math style="vertical-align: -5px">f(x)</math> is continuous at <math style="vertical-align: 0px">x=a</math> if
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| − | |-
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| − | | <math style="vertical-align: -14px">\lim_{x\rightarrow a^+}f(x)=\lim_{x\rightarrow a^-}f(x)=f(a).</math>
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| − | |}
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| | + | [[009A Sample Final 3, Problem 4 Detailed Solution|'''<u>Detailed Solution</u>''']] |
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| − | '''Solution:'''
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |-
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| − | |We first calculate <math style="vertical-align: -14px">\lim_{x\rightarrow 0^+}f(x).</math> We have
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{\lim_{x\rightarrow 0^+}f(x)} & = & \displaystyle{\lim_{x\rightarrow 0^+} x-\cos x}\\
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| − | &&\\
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| − | & = & \displaystyle{0-\cos(0)}\\
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| − | &&\\
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| − | & = & \displaystyle{-1.}
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| − | \end{array}</math>
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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| − | |-
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| − | |Now, we calculate <math style="vertical-align: -14px">\lim_{x\rightarrow 0^-}f(x).</math> We have
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{\lim_{x\rightarrow 0^-}f(x)} & = & \displaystyle{\lim_{x\rightarrow 0^-} \frac{x}{|x|}}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{x\rightarrow 0^-} \frac{x}{-x}}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{x\rightarrow 0^-} -1}\\
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| − | &&\\
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| − | & = & \displaystyle{-1.}
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| − | \end{array}</math>
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 3:
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| − | |-
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| − | |Since
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| − | <math style="vertical-align: -15px">\lim_{x\rightarrow 3^+}f(x)=\lim_{x\rightarrow 3^-}f(x)=-1,</math>
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| − | |-
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| − | |we have
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| − | | <math>\lim_{x\rightarrow 3} f(x)=-1.</math>
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| − | |-
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| − | |But,
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| − | |-
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| − | | <math>f(0)=0\ne \lim_{x\rightarrow 3} f(x).</math>
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| − | |-
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| − | |Thus, <math style="vertical-align: -5px">f(x)</math> is not continuous.
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| − | |-
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| − | |It is a jump discontinuity.
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Final Answer:
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| − | | <math style="vertical-align: -5px">f(x)</math> is not continuous. It is a jump discontinuity.
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| − | |}
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| | [[009A_Sample_Final_3|'''<u>Return to Sample Exam</u>''']] | | [[009A_Sample_Final_3|'''<u>Return to Sample Exam</u>''']] |
