009A Sample Midterm 1, Problem 2 - Revision history

MathAdmin at 21:40, 8 November 2017

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MathAdmin at 16:45, 7 November 2017

2017-11-07T16:45:46Z

MathAdmin at 15:28, 7 November 2017

2017-11-07T15:28:21Z

MathAdmin at 15:27, 7 November 2017

2017-11-07T15:27:56Z

MathAdmin: Created page with "Consider the following function $f:$ :: $f(x) = \left\{ \begin{array}{lr} x^2 & \text{if }x..."$

2017-04-09T18:34:12Z

Created page with "Consider the following function <math style="vertical-align: -5px"> f:</math> ::<math>f(x) = \left\{ \begin{array}{lr} x^2 & \text{if }x..."

New page

Consider the following function  <math style="vertical-align: -5px"> f:</math>
::<math>f(x) = \left\{
\begin{array}{lr}
x^2 & \text{if }x < 1\\
\sqrt{x} & \text{if }x \geq 1
\end{array}
\right.
</math>

(a) Find  <math style="vertical-align: -15px"> \lim_{x\rightarrow 1^-} f(x).</math>

(b) Find  <math style="vertical-align: -15px"> \lim_{x\rightarrow 1^+} f(x).</math>

(c) Find  <math style="vertical-align: -13px"> \lim_{x\rightarrow 1} f(x).</math>

(d) Is  <math style="vertical-align: -5px">f</math>  continuous at  <math style="vertical-align: -1px">x=1?</math>  Briefly explain.

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|'''1.''' If  <math style="vertical-align: -15px">\lim_{x\rightarrow a^-} f(x)=\lim_{x\rightarrow a^+} f(x)=c,</math>
|-
|        then  <math style="vertical-align: -12px">\lim_{x\rightarrow a} f(x)=c.</math>
|-
|'''2.'''  <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:'''

'''(a)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Notice that we are calculating a left hand limit.
|-
|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>
|-
|Using the definition of  <math style="vertical-align: -5px">f(x),</math>  we have
|-
|        <math>\lim_{x\rightarrow 1^-} f(x)=\lim_{x\rightarrow 1^-} x^2.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we have
|-
|
        <math>\begin{array}{rcl}
\displaystyle{\lim_{x\rightarrow 1^-} f(x)} & = & \displaystyle{\lim_{x\rightarrow 1^-} x^2}\\
&&\\
& = & \displaystyle{\lim_{x\rightarrow 1} x^2}\\
&&\\
& = & \displaystyle{1^2}\\
&&\\
& = & \displaystyle{1.}\\
\end{array}</math>
|}

'''(b)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Notice that we are calculating a right hand limit.
|-
|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>
|-
|Using the definition of  <math style="vertical-align: -5px">f(x),</math>  we have
|-
|        <math>\lim_{x\rightarrow 1^+} f(x)=\lim_{x\rightarrow 1^+} \sqrt{x}.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we have
|-
|
        <math>\begin{array}{rcl}
\displaystyle{\lim_{x\rightarrow 1^+} f(x)} & = & \displaystyle{\lim_{x\rightarrow 1^+} \sqrt{x}}\\
&&\\
& = & \displaystyle{\lim_{x\rightarrow 1} \sqrt{x}}\\
&&\\
& = & \displaystyle{\sqrt{1}}\\
&&\\
& = & \displaystyle{1.}\\
\end{array}</math>
|}

'''(c)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|From (a) and (b), we have
|-
|        <math>\lim_{x\rightarrow 1^-}f(x)=1</math>
|-
|and
|-
|        <math>\lim_{x\rightarrow 1^+}f(x)=1.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Since
|-
|        <math>\lim_{x\rightarrow 1^-}f(x)=\lim_{x\rightarrow 1^+}f(x)=1,</math>
|-
|we have
|-
|        <math>\lim_{x\rightarrow 1}f(x)=1.</math>
|}

'''(d)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|From (c), we have
|-
|        <math>\lim_{x\rightarrow 1}f(x)=1.</math>
|-
|Also,
|-
|        <math>f(1)=\sqrt{1}=1.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Since
|-
|        <math>\lim_{x\rightarrow 1}f(x)=f(1),</math>
|-
|<math style="vertical-align: -5px">f(x)</math>  is continuous at  <math style="vertical-align: -1px">x=1.</math>
|-
|
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|    '''(a)'''     <math>1</math>
|-
|    '''(b)'''     <math>1</math>
|-
|    '''(c)'''     <math>1</math>
|-
|    '''(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>
|}
[[009A_Sample_Midterm_1|'''Return to Sample Exam''']]

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 <span class="exam">(b) Show that at time &nbsp;<math style="vertical-align: -4px">t=5,</math>&nbsp; the size of the population is half its limiting size.
 <hr>
-[[009A Sample Midterm 1, Problem 2 Detailed Solution|'''<u>Detailed Solution with Background Information</u>''']]
+[[009A Sample Midterm 1, Problem 2 Solution|'''<u>Solution</u>''']]
 [[009A_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 6: / Line 6: @@
 <span class="exam">(b) Show that at time &nbsp;<math style="vertical-align: -4px">t=5,</math>&nbsp; the size of the population is half its limiting size.
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+[[File:9ASM1P2.jpg|600px|thumb|center]]
 [[009A_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

← Older revision		Revision as of 15:28, 7 November 2017
Line 3:		Line 3:
	::<math>N(t)=\frac{1000t}{5+t},~t\ge 0.</math>		::<math>N(t)=\frac{1000t}{5+t},~t\ge 0.</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.

		+	<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.

	{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"		{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"