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| − | <span class="exam">Consider the following function <math style="vertical-align: -5px"> f:</math> | + | <span class="exam">Suppose the size of a population at time <math style="vertical-align: 0px">t</math> is given by |
| − | ::<math>f(x) = \left\{
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| − | \begin{array}{lr}
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| − | x^2 & \text{if }x < 1\\
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| − | \sqrt{x} & \text{if }x \geq 1
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| − | \end{array}
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| − | \right.
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| − | </math>
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| | | | |
| − | <span class="exam">(a) Find <math style="vertical-align: -15px"> \lim_{x\rightarrow 1^-} f(x).</math> | + | ::<math>N(t)=\frac{1000t}{5+t},~t\ge 0.</math> |
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| − | <span class="exam">(b) Find <math style="vertical-align: -15px"> \lim_{x\rightarrow 1^+} f(x).</math> | + | <span class="exam">(a) Determine the size of the population as <math style="vertical-align: -1px">t\rightarrow \infty.</math> We call this the limiting population size. |
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| − | <span class="exam">(c) Find <math style="vertical-align: -13px"> \lim_{x\rightarrow 1} f(x).</math> | + | <span class="exam">(b) Show that at time <math style="vertical-align: -4px">t=5,</math> the size of the population is half its limiting size. |
| | + | <hr> |
| | + | [[009A Sample Midterm 1, Problem 2 Solution|'''<u>Solution</u>''']] |
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| − | <span class="exam">(d) Is <math style="vertical-align: -5px">f</math> continuous at <math style="vertical-align: -1px">x=1?</math> Briefly explain.
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| | + | [[009A Sample Midterm 1, Problem 2 Detailed Solution|'''<u>Detailed 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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| − | |'''1.''' If <math style="vertical-align: -15px">\lim_{x\rightarrow a^-} f(x)=\lim_{x\rightarrow a^+} f(x)=c,</math>
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| − | |-
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| − | | then <math style="vertical-align: -12px">\lim_{x\rightarrow a} f(x)=c.</math>
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| − | |-
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| − | |'''2.''' <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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| − |
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| − |
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| − | '''Solution:'''
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| − |
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| − | '''(a)'''
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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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| − | |Notice that we are calculating a left hand limit.
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| − | |-
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| − | |Thus, we are looking at values of <math style="vertical-align: 0px">x</math> that are smaller than <math style="vertical-align: -1px">1.</math>
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| − | |-
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| − | |Using the definition of <math style="vertical-align: -5px">f(x),</math> we have
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| − | |-
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| − | | <math>\lim_{x\rightarrow 1^-} f(x)=\lim_{x\rightarrow 1^-} x^2.</math>
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| − | |}
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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 have
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| − | |-
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| − | |
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{\lim_{x\rightarrow 1^-} f(x)} & = & \displaystyle{\lim_{x\rightarrow 1^-} x^2}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{x\rightarrow 1} x^2}\\
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| − | &&\\
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| − | & = & \displaystyle{1^2}\\
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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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| − |
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| − | '''(b)'''
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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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| − | |Notice that we are calculating a right hand limit.
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| − | |-
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| − | |Thus, we are looking at values of <math style="vertical-align: 0px">x</math> that are bigger than <math style="vertical-align: -2px">1.</math>
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| − | |-
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| − | |Using the definition of <math style="vertical-align: -5px">f(x),</math> we have
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| − | |-
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| − | | <math>\lim_{x\rightarrow 1^+} f(x)=\lim_{x\rightarrow 1^+} \sqrt{x}.</math>
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| − | |}
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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 have
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| − | |-
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| − | |
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{\lim_{x\rightarrow 1^+} f(x)} & = & \displaystyle{\lim_{x\rightarrow 1^+} \sqrt{x}}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{x\rightarrow 1} \sqrt{x}}\\
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| − | &&\\
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| − | & = & \displaystyle{\sqrt{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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| − |
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| − | '''(c)'''
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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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| − | |From (a) and (b), we have
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| − | |-
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| − | | <math>\lim_{x\rightarrow 1^-}f(x)=1</math>
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| − | |-
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| − | |and
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| − | |-
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| − | | <math>\lim_{x\rightarrow 1^+}f(x)=1.</math>
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| − | |}
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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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| − | |Since
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| − | |-
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| − | | <math>\lim_{x\rightarrow 1^-}f(x)=\lim_{x\rightarrow 1^+}f(x)=1,</math>
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| − | |-
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| − | |we have
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| − | |-
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| − | | <math>\lim_{x\rightarrow 1}f(x)=1.</math>
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| − | |}
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| − |
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| − | '''(d)'''
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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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| − | |From (c), we have
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| − | |-
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| − | | <math>\lim_{x\rightarrow 1}f(x)=1.</math>
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| − | |-
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| − | |Also,
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| − | |-
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| − | | <math>f(1)=\sqrt{1}=1.</math>
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| − | |}
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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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| − | |Since
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| − | |-
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| − | | <math>\lim_{x\rightarrow 1}f(x)=f(1),</math>
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| − | |-
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| − | |<math style="vertical-align: -5px">f(x)</math> is continuous at <math style="vertical-align: -1px">x=1.</math>
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| − | |-
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| − | |
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| − | |}
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| − |
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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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| − | |-
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| − | | '''(a)''' <math>1</math>
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| − | |-
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| − | | '''(b)''' <math>1</math>
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| − | |-
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| − | | '''(c)''' <math>1</math>
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| − | |-
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| − | | '''(d)''' <math style="vertical-align: -5px">f(x)</math> is continuous at <math style="vertical-align: -1px">x=1</math> since <math style="vertical-align: -12px">\lim_{x\rightarrow 1}f(x)=f(1).</math>
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| − | |}
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| | [[009A_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']] | | [[009A_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']] |
