009C Sample Final 1, Problem 1 - Revision history

Kayla Murray at 00:55, 3 December 2017

2017-12-03T00:55:12Z

2017-03-18T18:48:14Z

2017-02-26T22:54:42Z

2017-02-26T22:50:19Z

2017-02-25T23:45:20Z

2017-02-25T23:43:01Z

2017-02-25T23:41:33Z

2017-02-25T23:40:52Z

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2017-02-25T23:39:36Z

@@ Line 5: / Line 5: @@
 <span class="exam">(b) &nbsp; <math style="vertical-align: -12px">\lim_{n\rightarrow \infty} \frac{\ln n}{\ln 3n}</math>
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+<hr>
-!Foundations: &nbsp;
+[[009C Sample Final 1, Problem 1 Solution|'''<u>Solution</u>''']]
-'''Solution:'''
+[[009C Sample Final 1, Problem 1 Detailed Solution|'''<u>Detailed Solution</u>''']]
 [[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 38: / Line 38: @@
 & \overset{L'H}{=} & \displaystyle{\frac{-4}{10}}\\
 &&\\
-& = & \displaystyle{\frac{-2}{5}}.
+& = & \displaystyle{-\frac{2}{5}}.
 \end{array}</math>
 |}
@@ Line 48: / Line 48: @@
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp;<math>\lim_{n\rightarrow \infty} \frac{3-2n^2}{5n^2+n+1}=\frac{-2}{5}.</math>
+&nbsp; &nbsp; &nbsp; &nbsp;<math>\lim_{n\rightarrow \infty} \frac{3-2n^2}{5n^2+n+1}=-\frac{2}{5}.</math>
 |}
@@ Line 62: / Line 62: @@
+|
 &nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
-\displaystyle{\lim_{x \rightarrow \infty}\frac{\ln x}{\ln(3x)}} & \overset{l'H}{=} & \displaystyle{\lim_{x \rightarrow \infty}\frac{\frac{1}{x}}{\frac{3}{3x}}}\\
+\displaystyle{\lim_{x \rightarrow \infty}\frac{\ln x}{\ln(3x)}} & \overset{l'H}{=} & \displaystyle{\lim_{x \rightarrow \infty}\frac{\big(\frac{1}{x}\big)}{\big(\frac{3}{3x}\big)}}\\
 &&\\
 & = & \displaystyle{\lim_{x \rightarrow \infty} 1}\\
@@ Line 83: / Line 83: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp;&nbsp; '''(a)''' &nbsp; &nbsp; <math style="vertical-align: -14px">\frac{-2}{5}</math>
+|&nbsp;&nbsp; '''(a)''' &nbsp; &nbsp; <math style="vertical-align: -14px">-\frac{2}{5}</math>
 |-
 |&nbsp;&nbsp; '''(b)''' &nbsp; &nbsp; <math style="vertical-align: -3px">1</math>
 |}
 [[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 11: / Line 11: @@
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp; Suppose that <math>\lim_{x\rightarrow \infty} f(x)</math> and <math>\lim_{x\rightarrow \infty} g(x)</math> are both zero or both <math style="vertical-align: -1px">\pm \infty .</math>
+&nbsp; &nbsp; &nbsp; &nbsp; Suppose that <math>\lim_{x\rightarrow \infty} f(x)</math> &nbsp; and <math>\lim_{x\rightarrow \infty} g(x)</math> &nbsp; are both zero or both &nbsp; <math style="vertical-align: -1px">\pm \infty .</math>
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp;If <math>\lim_{x\rightarrow \infty} \frac{f'(x)}{g'(x)}</math> is finite or <math style="vertical-align: -1px">\pm \infty ,</math>
+&nbsp; &nbsp; &nbsp; &nbsp;If <math>\lim_{x\rightarrow \infty} \frac{f'(x)}{g'(x)}</math> &nbsp; is finite or &nbsp; <math style="vertical-align: -4px">\pm \infty ,</math>
 |-
+|

@@ Line 1: / Line 1: @@
 <span class="exam">Compute
-<span class="exam">(a) <math style="vertical-align: -12px">\lim_{n\rightarrow \infty} \frac{3-2n^2}{5n^2+n+1}</math>
+<span class="exam">(a) &nbsp; <math style="vertical-align: -12px">\lim_{n\rightarrow \infty} \frac{3-2n^2}{5n^2+n+1}</math>
-<span class="exam">(b) <math style="vertical-align: -12px">\lim_{n\rightarrow \infty} \frac{\ln n}{\ln 3n}</math>
+<span class="exam">(b) &nbsp; <math style="vertical-align: -12px">\lim_{n\rightarrow \infty} \frac{\ln n}{\ln 3n}</math>
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"

@@ Line 19: / Line 19: @@
 &nbsp; &nbsp; &nbsp; &nbsp;then <math>\lim_{x\rightarrow \infty} \frac{f(x)}{g(x)}=\lim_{x\rightarrow \infty} \frac{f'(x)}{g'(x)}.</math>
 |}
 '''Solution:'''
@@ Line 77: / Line 78: @@
 &nbsp; &nbsp; &nbsp; &nbsp;<math>\lim_{n\rightarrow \infty} \frac{\ln n}{\ln 3n}=1.</math>
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"

@@ Line 32: / Line 32: @@
 |-
+|
-::<math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 \displaystyle{\lim_{x \rightarrow \infty}\frac{3-2x^2}{5x^2 + x +1}} & \overset{L'H}{=} & \displaystyle{\lim_{x \rightarrow \infty}\frac{-4x}{10x+1}}\\
 &&\\
@@ Line 47: / Line 47: @@
 |-
+|
-::<math>\lim_{n\rightarrow \infty} \frac{3-2n^2}{5n^2+n+1}=\frac{-2}{5}.</math>
+&nbsp; &nbsp; &nbsp; &nbsp;<math>\lim_{n\rightarrow \infty} \frac{3-2n^2}{5n^2+n+1}=\frac{-2}{5}.</math>
 |}
@@ Line 60: / Line 60: @@
 |-
+|
-::<math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 \displaystyle{\lim_{x \rightarrow \infty}\frac{\ln x}{\ln(3x)}} & \overset{l'H}{=} & \displaystyle{\lim_{x \rightarrow \infty}\frac{\frac{1}{x}}{\frac{3}{3x}}}\\
 &&\\
@@ Line 75: / Line 75: @@
 |-
+|
-::<math>\lim_{n\rightarrow \infty} \frac{\ln n}{\ln 3n}=1.</math>
+&nbsp; &nbsp; &nbsp; &nbsp;<math>\lim_{n\rightarrow \infty} \frac{\ln n}{\ln 3n}=1.</math>
 |}
@@ Line 81: / Line 81: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp;&nbsp; '''(a)''' <math style="vertical-align: -14px">\frac{-2}{5}</math>
+|&nbsp;&nbsp; '''(a)''' &nbsp; &nbsp; <math style="vertical-align: -14px">\frac{-2}{5}</math>
 |-
-|&nbsp;&nbsp; '''(b)''' <math style="vertical-align: -3px">1</math>
+|&nbsp;&nbsp; '''(b)''' &nbsp; &nbsp; <math style="vertical-align: -3px">1</math>
 |}
 [[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 8: / Line 8: @@
 !Foundations: &nbsp;
 |-
 |'''L'Hopital's Rule'''
 |-
+|
-::Suppose that <math>\lim_{x\rightarrow \infty} f(x)</math> and <math>\lim_{x\rightarrow \infty} g(x)</math> are both zero or both <math style="vertical-align: -1px">\pm \infty .</math>
+&nbsp; &nbsp; &nbsp; &nbsp; Suppose that <math>\lim_{x\rightarrow \infty} f(x)</math> and <math>\lim_{x\rightarrow \infty} g(x)</math> are both zero or both <math style="vertical-align: -1px">\pm \infty .</math>
 |-
+|
-::If <math>\lim_{x\rightarrow \infty} \frac{f'(x)}{g'(x)}</math> is finite or <math style="vertical-align: -1px">\pm \infty ,</math>
+&nbsp; &nbsp; &nbsp; &nbsp;If <math>\lim_{x\rightarrow \infty} \frac{f'(x)}{g'(x)}</math> is finite or <math style="vertical-align: -1px">\pm \infty ,</math>
 |-
+|
-::then <math>\lim_{x\rightarrow \infty} \frac{f(x)}{g(x)}=\lim_{x\rightarrow \infty} \frac{f'(x)}{g'(x)}.</math>
+&nbsp; &nbsp; &nbsp; &nbsp;then <math>\lim_{x\rightarrow \infty} \frac{f(x)}{g(x)}=\lim_{x\rightarrow \infty} \frac{f'(x)}{g'(x)}.</math>
 |}