009A Sample Midterm 1, Problem 1 - Revision history

Kayla Murray at 01:36, 7 November 2017

2017-11-07T01:36:59Z

Kayla Murray at 01:36, 7 November 2017

2017-11-07T01:36:12Z

Kayla Murray at 18:23, 4 November 2017

2017-11-04T18:23:47Z

Kayla Murray at 14:03, 3 November 2017

2017-11-03T14:03:36Z

Kayla Murray at 17:43, 27 March 2017

2017-03-27T17:43:30Z

Kayla Murray at 17:03, 13 March 2017

2017-03-13T17:03:57Z

Kayla Murray at 23:43, 26 February 2017

2017-02-26T23:43:03Z

Kayla Murray at 23:33, 26 February 2017

2017-02-26T23:33:56Z

Kayla Murray at 23:35, 18 February 2017

2017-02-18T23:35:53Z

Kayla Murray at 23:27, 18 February 2017

2017-02-18T23:27:41Z

@@ Line 9: / Line 9: @@
 [[009A Sample Midterm 1, Problem 1 Detailed Solution|'''<u>Detailed Solution with Background Information</u>''']]
-[[File:9ASM1P3.jpg|600px|thumb|center]]
+[[File:9ASM1P1.jpg|600px|thumb|center]]
 [[009A_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 7: / Line 7: @@
 <span class="exam">(c) Evaluate &nbsp;<math style="vertical-align: -14px">\lim _{x\rightarrow -3^+} \frac{x}{x^2-9} </math>
 <hr>
-<pdf>File:9ASM1P1HW.pdf</pdf>
+[[009A Sample Midterm 1, Problem 1 Detailed Solution|'''<u>Detailed Solution with Background Information</u>''']]
-[[009A Sample Midterm 1, Problem 1 Detailed Solution|'''<u>Detailed Solution for this Problem</u>''']]
+[[File:9ASM1P3.jpg|600px|thumb|center]]
 [[009A_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 7: / Line 7: @@
 <span class="exam">(c) Evaluate &nbsp;<math style="vertical-align: -14px">\lim _{x\rightarrow -3^+} \frac{x}{x^2-9} </math>
 <hr>
-(insert picture of handwritten solution)
+<pdf>File:9ASM1P1HW.pdf</pdf>
 [[009A Sample Midterm 1, Problem 1 Detailed Solution|'''<u>Detailed Solution for this Problem</u>''']]
 [[009A_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

← Older revision		Revision as of 14:03, 3 November 2017
Line 6:		Line 6:

	<span class="exam">(c) Evaluate  <math style="vertical-align: -14px">\lim _{x\rightarrow -3^+} \frac{x}{x^2-9} </math>		<span class="exam">(c) Evaluate  <math style="vertical-align: -14px">\lim _{x\rightarrow -3^+} \frac{x}{x^2-9} </math>
		+	<hr>
		+	(insert picture of handwritten solution)

		+	[[009A Sample Midterm 1, Problem 1 Detailed Solution\|'''<u>Detailed Solution for this Problem</u>''']]

−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Foundations:  ~~
−	\|-
−	~~\| '''1.''' If  <math style="vertical-align: -12px">\lim_{x\rightarrow a} g(x)\neq 0,</math>  we have~~
−	\|-
−	~~\|        <math>\lim_{x\rightarrow a} \frac{f(x)}{g(x)}=\frac{\displaystyle{\lim_{x\rightarrow a} f(x)}}{\displaystyle{\lim_{x\rightarrow a} g(x)}}.</math>~~
−	\|-
−	~~\| '''2.''' Recall~~
−	\|-
−	~~\|       <math style="vertical-align: -14px">\lim_{x\rightarrow 0} \frac{\sin x}{x}=1</math>~~
−	\|}
−
−
−	~~'''Solution:'''~~
−
−	~~'''(a)'''~~
−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Step 1:  ~~
−	\|-
−	~~\|Since  <math style="vertical-align: -12px">\lim_{x\rightarrow 2} x =2\ne 0,</math>~~
−	\|-
−	~~\|we have~~
−	\|-
−	~~\|        <math>\begin{array}{rcl}~~
−	~~\displaystyle{5} & = & \displaystyle{\lim _{x\rightarrow 2} \bigg[\frac{4-g(x)}{x}\bigg]}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{\frac{\lim_{x\rightarrow 2} (4-g(x))}{\lim_{x\rightarrow 2} x}}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{\frac{\lim_{x\rightarrow 2} (4-g(x))}{2}.}~~
−	~~\end{array}</math>~~
−	\|}
−
−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Step 2:  ~~
−	\|-
−	~~\|If we multiply both sides of the last equation by  <math>2,</math>  we get~~
−	\|-
−	~~\|        <math>10=\lim_{x\rightarrow 2} (4-g(x)).</math>~~
−	\|-
−	~~\|Now, using linearity properties of limits, we have~~
−	\|-
−	~~\|        <math>\begin{array}{rcl}~~
−	~~\displaystyle{10} & = & \displaystyle{\lim_{x\rightarrow 2} 4 -\lim_{x\rightarrow 2}g(x)}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{4-\lim_{x\rightarrow 2} g(x).}\\~~
−	~~\end{array}</math>~~
−	\|}
−
−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Step 3:  ~~
−	\|-
−	~~\|Solving for  <math style="vertical-align: -12px">\lim_{x\rightarrow 2} g(x)</math>  in the last equation,~~
−	\|-
−	~~\|we get~~
−	\|-
−	\|
−	~~        <math> \lim_{x\rightarrow 2} g(x)=-6.</math>~~
−	\|}
−
−	~~'''(b)'''~~
−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Step 1:  ~~
−	\|-
−	~~\|First, we write~~
−	\|-
−	~~\|        <math>\lim_{x\rightarrow 0} \frac{\sin(4x)}{5x}=\lim_{x\rightarrow 0} \frac{4}{5} \bigg(\frac{\sin(4x)}{4x}\bigg).</math>~~
−	\|}
−
−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Step 2:  ~~
−	\|-
−	~~\|Now, we have~~
−	\|-
−	~~\|        <math>\begin{array}{rcl}~~
−	~~\displaystyle{\lim_{x\rightarrow 0} \frac{\sin(4x)}{5x}} & = & \displaystyle{\frac{4}{5}\lim_{x\rightarrow 0} \frac{\sin(4x)}{4x}}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{\frac{4}{5}(1)}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{\frac{4}{5}.}~~
−	~~\end{array}</math>~~
−	\|}
−
−	~~'''(c)'''~~
−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Step 1:  ~~
−	\|-
−	~~\|When we plug in  <math style="vertical-align: 0px">-3</math>  into   <math style="vertical-align: -12px">\frac{x}{x^2-9},</math>~~
−	\|-
−	~~\|we get   <math style="vertical-align: -12px">\frac{-3}{0}.</math>~~
−	\|-
−	~~\|Thus,~~
−	\|-
−	~~\|        <math>\lim_{x\rightarrow -3^+} \frac{x}{x^2-9}</math>~~
−	\|-
−	~~\|is either equal to  <math style="vertical-align: -1px">\infty</math>  or  <math style="vertical-align: -1px">-\infty.</math>~~
−	\|}
−
−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Step 2:  ~~
−	\|-
−	~~\|To figure out which one, we factor the denominator to get~~
−	\|-
−	~~\|        <math>\lim_{x\rightarrow -3^+} \frac{x}{x^2-9}=\lim_{x\rightarrow -3^+} \frac{x}{(x-3)(x+3)}.</math>~~
−	\|-
−	~~\|We are taking a right hand limit. So, we are looking at values of  <math style="vertical-align: 0px">x</math>~~
−	\|-
−	~~\|a little bigger than  <math style="vertical-align: 0px">-3.</math>  (You can imagine values like  <math style="vertical-align: 0px">x=-2.9.</math> )~~
−	\|-
−	~~\|For these values, the numerator will be negative.~~
−	\|-
−	~~\|Also, for these values,  <math style="vertical-align: 0px">x-3</math>  will be negative and  <math style="vertical-align: -1px">x+3</math>  will be positive.~~
−	\|-
−	~~\|Therefore, the denominator will be negative.~~
−	\|-
−	~~\|Since both the numerator and denominator will be negative (have the same sign),~~
−	\|-
−	~~\|        <math>\lim_{x\rightarrow -3^+} \frac{x}{x^2-9}=\infty.</math>~~
−	\|}
−
−
−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Final Answer:  ~~
−	\|-
−	~~\|    '''(a)'''     <math> -6</math>~~
−	\|-
−	~~\|    '''(b)'''     <math>\frac{4}{5}</math>~~
−	\|-
−	~~\|    '''(c)'''     <math>\infty</math>~~
−	\|}
	[[009A_Sample_Midterm_1\|'''<u>Return to Sample Exam</u>''']]		[[009A_Sample_Midterm_1\|'''<u>Return to Sample Exam</u>''']]

@@ Line 15: / Line 15: @@
 |&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{x\rightarrow a} \frac{f(x)}{g(x)}=\frac{\displaystyle{\lim_{x\rightarrow a} f(x)}}{\displaystyle{\lim_{x\rightarrow a} g(x)}}.</math>
 |-
-| '''2.''' &nbsp;<math style="vertical-align: -14px">\lim_{x\rightarrow 0} \frac{\sin x}{x}=1</math>
+| '''2.''' Recall
 |}

@@ Line 100: / Line 100: @@
 |&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{x\rightarrow -3^+} \frac{x}{x^2-9}</math>
 |-
-|is either equal to &nbsp;<math style="vertical-align: -1px">+\infty</math>&nbsp; or &nbsp;<math style="vertical-align: -1px">-\infty.</math>
+|is either equal to &nbsp;<math style="vertical-align: -1px">\infty</math>&nbsp; or &nbsp;<math style="vertical-align: -1px">-\infty.</math>
 |}
@@ Line 122: / Line 122: @@
 |Since both the numerator and denominator will be negative (have the same sign),
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{x\rightarrow -3^+} \frac{x}{x^2-9}=+\infty.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{x\rightarrow -3^+} \frac{x}{x^2-9}=\infty.</math>
 |}
@@ Line 129: / Line 129: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; <math> \lim_{x\rightarrow 2} g(x)=-6</math>
+|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; <math> -6</math>
 |-
 |&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; <math>\frac{4}{5}</math>
 |-
-|&nbsp; &nbsp; '''(c)''' &nbsp; &nbsp; <math>+\infty</math>
+|&nbsp; &nbsp; '''(c)''' &nbsp; &nbsp; <math>\infty</math>
 |}
 [[009A_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 11: / Line 11: @@
 !Foundations: &nbsp;
 |-
-| '''1.''' If <math style="vertical-align: -12px">\lim_{x\rightarrow a} g(x)\neq 0,</math> we have
+| '''1.''' If &nbsp;<math style="vertical-align: -12px">\lim_{x\rightarrow a} g(x)\neq 0,</math>&nbsp; we have
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{x\rightarrow a} \frac{f(x)}{g(x)}=\frac{\displaystyle{\lim_{x\rightarrow a} f(x)}}{\displaystyle{\lim_{x\rightarrow a} g(x)}}.</math>
 |-
-| '''2.''' <math style="vertical-align: -14px">\lim_{x\rightarrow 0} \frac{\sin x}{x}=1</math>
+| '''2.''' &nbsp;<math style="vertical-align: -14px">\lim_{x\rightarrow 0} \frac{\sin x}{x}=1</math>
 |}
@@ Line 25: / Line 25: @@
 !Step 1: &nbsp;
 |-
-|Since <math style="vertical-align: -12px">\lim_{x\rightarrow 2} x =2\ne 0,</math>
+|Since &nbsp;<math style="vertical-align: -12px">\lim_{x\rightarrow 2} x =2\ne 0,</math>
 |-
 |we have
@@ Line 41: / Line 41: @@
 !Step 2: &nbsp;
 |-
-|If we multiply both sides of the last equation by <math>2,</math> we get
+|If we multiply both sides of the last equation by &nbsp;<math>2,</math>&nbsp; we get
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp; <math>10=\lim_{x\rightarrow 2} (4-g(x)).</math>
@@ Line 57: / Line 57: @@
 !Step 3: &nbsp;
 |-
-|Solving for <math style="vertical-align: -12px">\lim_{x\rightarrow 2} g(x)</math> in the last equation,
+|Solving for &nbsp;<math style="vertical-align: -12px">\lim_{x\rightarrow 2} g(x)</math>&nbsp; in the last equation,
 |-
 |we get
@@ Line 92: / Line 92: @@
 !Step 1: &nbsp;
 |-
-|When we plug in <math style="vertical-align: 0px">-3</math> into &nbsp; <math style="vertical-align: -12px">\frac{x}{x^2-9},</math>
+|When we plug in &nbsp;<math style="vertical-align: 0px">-3</math>&nbsp; into &nbsp; <math style="vertical-align: -12px">\frac{x}{x^2-9},</math>
 |-
 |we get &nbsp; <math style="vertical-align: -12px">\frac{-3}{0}.</math>
@@ Line 100: / Line 100: @@
 |&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{x\rightarrow -3^+} \frac{x}{x^2-9}</math>
 |-
-|is either equal to <math style="vertical-align: -1px">+\infty</math> or <math style="vertical-align: -1px">-\infty.</math>
+|is either equal to &nbsp;<math style="vertical-align: -1px">+\infty</math>&nbsp; or &nbsp;<math style="vertical-align: -1px">-\infty.</math>
 |}
@@ Line 110: / Line 110: @@
 |&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{x\rightarrow -3^+} \frac{x}{x^2-9}=\lim_{x\rightarrow -3^+} \frac{x}{(x-3)(x+3)}.</math>
 |-
-|We are taking a right hand limit. So, we are looking at values of <math style="vertical-align: 0px">x</math>
+|We are taking a right hand limit. So, we are looking at values of &nbsp;<math style="vertical-align: 0px">x</math>
 |-
-|a little bigger than <math style="vertical-align: 0px">-3.</math> (You can imagine values like <math style="vertical-align: 0px">x=-2.9.</math>)
+|a little bigger than &nbsp;<math style="vertical-align: 0px">-3.</math>&nbsp; (You can imagine values like &nbsp;<math style="vertical-align: 0px">x=-2.9.</math>&nbsp;)
 |-
 |For these values, the numerator will be negative.
 |-
-|Also, for these values, <math style="vertical-align: 0px">x-3</math> will be negative and <math style="vertical-align: -1px">x+3</math> will be positive.
+|Also, for these values, &nbsp;<math style="vertical-align: 0px">x-3</math>&nbsp; will be negative and &nbsp;<math style="vertical-align: -1px">x+3</math>&nbsp; will be positive.
 |-
 |Therefore, the denominator will be negative.

@@ Line 1: / Line 1: @@
 <span class="exam">Find the following limits:
-<span class="exam">(a) Find <math style="vertical-align: -13px">\lim _{x\rightarrow 2} g(x),</math> provided that <math style="vertical-align: -15px">\lim _{x\rightarrow 2} \bigg[\frac{4-g(x)}{x}\bigg]=5</math>
+<span class="exam">(a) Find &nbsp;<math style="vertical-align: -13px">\lim _{x\rightarrow 2} g(x),</math>&nbsp; provided that &nbsp;<math style="vertical-align: -15px">\lim _{x\rightarrow 2} \bigg[\frac{4-g(x)}{x}\bigg]=5.</math>
-<span class="exam">(b) Find <math style="vertical-align: -14px">\lim _{x\rightarrow 0} \frac{\sin(4x)}{5x} </math>
+<span class="exam">(b) Find &nbsp;<math style="vertical-align: -14px">\lim _{x\rightarrow 0} \frac{\sin(4x)}{5x} </math>
-<span class="exam">(c) Evaluate <math style="vertical-align: -14px">\lim _{x\rightarrow -3^+} \frac{x}{x^2-9} </math>
+<span class="exam">(c) Evaluate &nbsp;<math style="vertical-align: -14px">\lim _{x\rightarrow -3^+} \frac{x}{x^2-9} </math>