009A Sample Midterm 1, Problem 2 - Revision history

Kayla Murray at 01:38, 7 November 2017

2017-11-07T01:38:44Z

Kayla Murray at 14:59, 3 November 2017

2017-11-03T14:59:46Z

Kayla Murray at 18:20, 18 March 2017

2017-03-18T18:20:26Z

Kayla Murray at 17:06, 13 March 2017

2017-03-13T17:06:19Z

Kayla Murray at 23:46, 26 February 2017

2017-02-26T23:46:48Z

Kayla Murray at 23:36, 26 February 2017

2017-02-26T23:36:56Z

Kayla Murray at 23:46, 18 February 2017

2017-02-18T23:46:18Z

Kayla Murray at 23:41, 18 February 2017

2017-02-18T23:41:35Z

Kayla Murray at 23:39, 18 February 2017

2017-02-18T23:39:55Z

Kayla Murray at 21:56, 18 February 2017

2017-02-18T21:56:41Z

@@ Line 7: / Line 7: @@
 <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>
-(insert picture of handwritten solution)
+[[File:9ASM1P2.jpg|600px|thumb|center]]
 [[009A_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 1: / Line 1: @@
-<span class="exam">Consider the following function &nbsp;<math style="vertical-align: -5px"> f:</math>
+<span class="exam">Suppose the size of a population at time &nbsp;<math style="vertical-align: 0px">t</math>&nbsp; is given by
-<span class="exam">(a) Find &nbsp;<math style="vertical-align: -15px"> \lim_{x\rightarrow 1^-} f(x).</math>
+::<math>N(t)=\frac{1000t}{5+t},~t\ge 0.</math>
-<span class="exam">(b) Find &nbsp;<math style="vertical-align: -15px"> \lim_{x\rightarrow 1^+} f(x).</math>
+<span class="exam">(a) Determine the size of the population as &nbsp;<math style="vertical-align: -1px">t\rightarrow \infty.</math>&nbsp; We call this the limiting population size.
-<span class="exam">(c) Find &nbsp;<math style="vertical-align: -13px"> \lim_{x\rightarrow 1} f(x).</math>
+<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.
-<span class="exam">(d) Is &nbsp;<math style="vertical-align: -5px">f</math>&nbsp; continuous at &nbsp;<math style="vertical-align: -1px">x=1?</math>&nbsp; Briefly explain.
+(insert picture of handwritten solution)
 [[009A_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 153: / Line 153: @@
 |&nbsp; &nbsp; '''(c)''' &nbsp; &nbsp; <math>1</math>
 |-
-|&nbsp; &nbsp; '''(d)''' &nbsp; &nbsp; <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>
+|&nbsp; &nbsp; '''(d)''' &nbsp; &nbsp; <math style="vertical-align: -5px">f(x)</math>&nbsp; is continuous at &nbsp;<math style="vertical-align: -1px">x=1</math>&nbsp; since &nbsp;<math style="vertical-align: -12px">\lim_{x\rightarrow 1}f(x)=f(1).</math>
 |}
 [[009A_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 39: / Line 39: @@
 |Notice that we are calculating a left hand limit.
 |-
-|Thus, we are looking at values of &nbsp;<math style="vertical-align: 0px">x</math>&nbsp; that are smaller than &nbsp;<math style="vertical-align: -2px">1.</math>
+|Thus, we are looking at values of &nbsp;<math style="vertical-align: 0px">x</math>&nbsp; that are smaller than &nbsp;<math style="vertical-align: -1px">1.</math>
 |-
 |Using the definition of &nbsp;<math style="vertical-align: -5px">f(x),</math>&nbsp; we have

@@ Line 21: / Line 21: @@
 !Foundations: &nbsp;
 |-
-|'''1.''' If <math style="vertical-align: -15px">\lim_{x\rightarrow a^-} f(x)=\lim_{x\rightarrow a^+} f(x)=c,</math>
+|'''1.''' If &nbsp;<math style="vertical-align: -15px">\lim_{x\rightarrow a^-} f(x)=\lim_{x\rightarrow a^+} f(x)=c,</math>
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; then <math style="vertical-align: -12px">\lim_{x\rightarrow a} f(x)=c.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; then &nbsp;<math style="vertical-align: -12px">\lim_{x\rightarrow a} f(x)=c.</math>
 |-
-|'''2.''' '''Definition of continuous'''
+|'''2.''' &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; is continuous at &nbsp;<math style="vertical-align: 0px">x=a</math>&nbsp; if
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -14px">\lim_{x\rightarrow a^+}f(x)=\lim_{x\rightarrow a^-}f(x)=f(a).</math>
@@ Line 41: / Line 39: @@
 |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: -2px">1.</math>
+|Thus, we are looking at values of &nbsp;<math style="vertical-align: 0px">x</math>&nbsp; that are smaller than &nbsp;<math style="vertical-align: -2px">1.</math>
 |-
-|Using the definition of <math style="vertical-align: -5px">f(x)</math>, we have
+|Using the definition of &nbsp;<math style="vertical-align: -5px">f(x),</math>&nbsp; we have
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{x\rightarrow 1^-} f(x)=\lim_{x\rightarrow 1^-} x^2.</math>
@@ Line 71: / Line 69: @@
 |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>
+|Thus, we are looking at values of &nbsp;<math style="vertical-align: 0px">x</math>&nbsp; that are bigger than &nbsp;<math style="vertical-align: -2px">1.</math>
 |-
-|Using the definition of <math style="vertical-align: -5px">f(x)</math>, we have
+|Using the definition of &nbsp;<math style="vertical-align: -5px">f(x),</math>&nbsp; we have
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{x\rightarrow 1^+} f(x)=\lim_{x\rightarrow 1^+} \sqrt{x}.</math>
@@ Line 140: / Line 138: @@
 |&nbsp; &nbsp; &nbsp; &nbsp; <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>
+|<math style="vertical-align: -5px">f(x)</math> &nbsp;is continuous at &nbsp;<math style="vertical-align: -1px">x=1.</math>
 |-
+|

@@ Line 1: / Line 1: @@
-<span class="exam">Consider the following function <math style="vertical-align: -5px"> f:</math>
+<span class="exam">Consider the following function &nbsp;<math style="vertical-align: -5px"> f:</math>
 ::<math>f(x) = \left\{
       \begin{array}{lr}
@@ Line 8: / Line 8: @@
 </math>
-<span class="exam">(a) Find <math style="vertical-align: -15px"> \lim_{x\rightarrow 1^-} f(x).</math>
+<span class="exam">(a) Find &nbsp;<math style="vertical-align: -15px"> \lim_{x\rightarrow 1^-} f(x).</math>
-<span class="exam">(b) Find <math style="vertical-align: -15px"> \lim_{x\rightarrow 1^+} f(x).</math>
+<span class="exam">(b) Find &nbsp;<math style="vertical-align: -15px"> \lim_{x\rightarrow 1^+} f(x).</math>
-<span class="exam">(c) Find <math style="vertical-align: -13px"> \lim_{x\rightarrow 1} f(x).</math>
+<span class="exam">(c) Find &nbsp;<math style="vertical-align: -13px"> \lim_{x\rightarrow 1} f(x).</math>
-<span class="exam">(d) Is <math style="vertical-align: -5px">f</math> continuous at <math style="vertical-align: -1px">x=1?</math> Briefly explain.
+<span class="exam">(d) Is &nbsp;<math style="vertical-align: -5px">f</math>&nbsp; continuous at &nbsp;<math style="vertical-align: -1px">x=1?</math>&nbsp; Briefly explain.