009A Sample Final 1, Problem 1 - Revision history

Kayla Murray at 22:38, 2 December 2017

2017-12-02T22:38:27Z

Kayla Murray at 22:37, 2 December 2017

2017-12-02T22:37:56Z

Kayla Murray at 23:33, 20 May 2017

2017-05-20T23:33:32Z

Kayla Murray at 20:09, 18 March 2017

2017-03-18T20:09:36Z

Kayla Murray at 16:09, 27 February 2017

2017-02-27T16:09:19Z

Kayla Murray at 00:34, 26 February 2017

2017-02-26T00:34:28Z

Kayla Murray at 00:25, 26 February 2017

2017-02-26T00:25:04Z

Kayla Murray at 01:46, 19 February 2017

2017-02-19T01:46:03Z

Kayla Murray at 01:42, 19 February 2017

2017-02-19T01:42:31Z

Grad at 18:18, 4 March 2016

2016-03-04T18:18:18Z

@@ Line 8: / Line 8: @@
 <hr>
-[[009A Sample Midterm 2, Problem 1 Solution|'''<u>Solution</u>''']]
+[[009A Sample Final 1, Problem 1 Solution|'''<u>Solution</u>''']]
-[[009A Sample Midterm 2, Problem 1 Detailed Solution|'''<u>Detailed Solution</u>''']]
+[[009A Sample Final 1, Problem 1 Detailed Solution|'''<u>Detailed Solution</u>''']]
-[[009A_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]
+[[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 7: / Line 7: @@
 <span class="exam">(c) &nbsp; <math style="vertical-align: -14px">\lim_{x\rightarrow -\infty} \frac{3x}{\sqrt{4x^2+x+5}}</math>
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+<hr>
-!Foundations: &nbsp;
+[[009A Sample Midterm 2, Problem 1 Solution|'''<u>Solution</u>''']]
-'''Solution:'''
+[[009A Sample Midterm 2, Problem 1 Detailed Solution|'''<u>Detailed Solution</u>''']]
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+[[009A_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 10: / Line 10: @@
 !Foundations: &nbsp;
 |-
-|'''L'Hôpital's Rule'''
+|'''L'Hôpital's Rule, Part 1'''
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; Suppose that &nbsp;<math style="vertical-align: -11px">\lim_{x\rightarrow \infty} f(x)</math>&nbsp; and &nbsp;<math style="vertical-align: -11px">\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;on an open interval &nbsp;<math style="vertical-align: 0px">I</math>&nbsp; containing &nbsp;<math style="vertical-align: -5px">c,</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">g'(x)\ne 0</math>&nbsp; on &nbsp;<math style="vertical-align: 0px">I</math>&nbsp; except possibly at &nbsp;<math style="vertical-align: 0px">c.</math>&nbsp;
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp;Then, &nbsp; <math style="vertical-align: -18px">\lim_{x\rightarrow c} \frac{f(x)}{g(x)}=\lim_{x\rightarrow c} \frac{f'(x)}{g'(x)}.</math>
 |}

@@ Line 46: / Line 46: @@
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp;<math>\lim_{x\rightarrow -3} \frac{x^3-9x}{6+2x}\,=\,\frac{(-3)(-3-3)}{2}\,=\,\frac{18}{2}\,=\,9.</math>
+&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 |}
@@ Line 67: / Line 73: @@
 !Step 2: &nbsp;
 |-
-|This limit is &nbsp; <math>+\infty.</math>
+|This limit is &nbsp; <math>\infty.</math>
 |}
@@ Line 94: / Line 100: @@
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp;<math>\lim_{x\rightarrow -\infty} \frac{3x}{\sqrt{4x^2+x+5}}\,=\,\lim_{x\rightarrow -\infty} \frac{-3}{\sqrt{4+\frac{1}{x}+\frac{5}{x^2}}}.</math>
+&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-|-
+\displaystyle{\lim_{x\rightarrow -\infty} \frac{3x}{\sqrt{4x^2+x+5}}} & = & \displaystyle{\lim_{x\rightarrow -\infty} \frac{-3}{\sqrt{4+\frac{1}{x}+\frac{5}{x^2}}}}\\
-|So, we have
+&&\\
-|-
+& = & \displaystyle{-\frac{3}{\sqrt{4}}}\\
+&&\\
-&nbsp; &nbsp; &nbsp; &nbsp;<math>\lim_{x\rightarrow -\infty} \frac{3x}{\sqrt{4x^2+x+5}}=\frac{-3}{\sqrt{4}}=\frac{-3}{2}.</math>
+& = & \displaystyle{-\frac{3}{2}.}
 |}
@@ Line 108: / Line 115: @@
 |&nbsp; &nbsp; '''(a)'''&nbsp; &nbsp; <math style="vertical-align: 0px">9</math>
 |-
-|&nbsp; &nbsp; '''(b)'''&nbsp; &nbsp; <math style="vertical-align: 0px">+\infty</math>
+|&nbsp; &nbsp; '''(b)'''&nbsp; &nbsp; <math style="vertical-align: 0px">\infty</math>
 |-
 |&nbsp; &nbsp; '''(c)'''&nbsp; &nbsp; <math style="vertical-align: -15px">-\frac{3}{2}</math>
 |}
 [[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 12: / Line 12: @@
 |'''L'Hôpital's Rule'''
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; Suppose that <math style="vertical-align: -11px">\lim_{x\rightarrow \infty} f(x)</math>&thinsp; and <math style="vertical-align: -11px">\lim_{x\rightarrow \infty} g(x)</math>&thinsp; are both zero or both <math style="vertical-align: -1px">\pm \infty .</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; Suppose that &nbsp;<math style="vertical-align: -11px">\lim_{x\rightarrow \infty} f(x)</math>&nbsp; and &nbsp;<math style="vertical-align: -11px">\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 style="vertical-align: -19px">\lim_{x\rightarrow \infty} \frac{f'(x)}{g'(x)}</math>&thinsp; is finite or&thinsp; <math style="vertical-align: -4px">\pm \infty ,</math>
+&nbsp; &nbsp; &nbsp; &nbsp; If &nbsp;<math style="vertical-align: -19px">\lim_{x\rightarrow \infty} \frac{f'(x)}{g'(x)}</math>&nbsp; is finite or &nbsp;<math style="vertical-align: -4px">\pm \infty ,</math>
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp; then <math style="vertical-align: -19px">\lim_{x\rightarrow \infty} \frac{f(x)}{g(x)}\,=\,\lim_{x\rightarrow \infty} \frac{f'(x)}{g'(x)}.</math>
+&nbsp; &nbsp; &nbsp; &nbsp; then &nbsp;<math style="vertical-align: -19px">\lim_{x\rightarrow \infty} \frac{f(x)}{g(x)}\,=\,\lim_{x\rightarrow \infty} \frac{f'(x)}{g'(x)}.</math>
 |}
@@ Line 34: / Line 34: @@
 &nbsp; &nbsp; &nbsp; &nbsp;<math>\lim_{x\rightarrow -3} \frac{x^3-9x}{6+2x}\,=\,\lim_{x\rightarrow -3}\frac{x(x-3)(x+3)}{2(x+3)}.</math>
 |-
-|So, we can cancel <math style="vertical-align: -2px">x+3</math>&thinsp; in the numerator and denominator. Thus, we have
+|So, we can cancel &nbsp;<math style="vertical-align: -2px">x+3</math>&nbsp; in the numerator and denominator. Thus, we have
 |-
+|
@@ Line 43: / Line 43: @@
 !Step 2: &nbsp;
 |-
-|Now, we can just plug in <math style="vertical-align: -1px">x=-3</math>&thinsp; to get
+|Now, we can just plug in &nbsp;<math style="vertical-align: -1px">x=-3</math>&nbsp; to get
 |-
+|
@@ Line 80: / Line 80: @@
 &nbsp; &nbsp; &nbsp; &nbsp;<math>\lim_{x\rightarrow -\infty} \frac{3x}{\sqrt{4x^2+x+5}}=\lim_{x\rightarrow -\infty} \frac{3x}{\sqrt{x^2(4+\frac{1}{x}+\frac{5}{x^2}})}.</math>
 |-
-|Since we are looking at the limit as <math style="vertical-align: 0px">x</math> goes to negative infinity, we have <math style="vertical-align: -2px">\sqrt{x^2}=-x.</math>
+|Since we are looking at the limit as &nbsp;<math style="vertical-align: 0px">x</math>&nbsp; goes to negative infinity, we have &nbsp;<math style="vertical-align: -2px">\sqrt{x^2}=-x.</math>
 |-
 |So, we have

@@ Line 1: / Line 1: @@
 <span class="exam">In each part, compute the limit. If the limit is infinite, be sure to specify positive or negative infinity.
-<span class="exam">(a) <math style="vertical-align: -14px">\lim_{x\rightarrow -3} \frac{x^3-9x}{6+2x}</math>
+<span class="exam">(a) &nbsp; <math style="vertical-align: -14px">\lim_{x\rightarrow -3} \frac{x^3-9x}{6+2x}</math>
-<span class="exam">(b) <math style="vertical-align: -14px">\lim_{x\rightarrow 0^+} \frac{\sin (2x)}{x^2}</math>
+<span class="exam">(b) &nbsp; <math style="vertical-align: -14px">\lim_{x\rightarrow 0^+} \frac{\sin (2x)}{x^2}</math>
-<span class="exam">(c) <math style="vertical-align: -14px">\lim_{x\rightarrow -\infty} \frac{3x}{\sqrt{4x^2+x+5}}</math>
+<span class="exam">(c) &nbsp; <math style="vertical-align: -14px">\lim_{x\rightarrow -\infty} \frac{3x}{\sqrt{4x^2+x+5}}</math>
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"

@@ Line 9: / Line 9: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Foundations: &nbsp;
 |-
 |'''L'Hôpital's Rule'''
 |-
-|Suppose that <math style="vertical-align: -11px">\lim_{x\rightarrow \infty} f(x)</math>&thinsp; and <math style="vertical-align: -11px">\lim_{x\rightarrow \infty} g(x)</math>&thinsp; are both zero or both <math style="vertical-align: -1px">\pm \infty .</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; Suppose that <math style="vertical-align: -11px">\lim_{x\rightarrow \infty} f(x)</math>&thinsp; and <math style="vertical-align: -11px">\lim_{x\rightarrow \infty} g(x)</math>&thinsp; are both zero or both <math style="vertical-align: -1px">\pm \infty .</math>
 |-
+|
-::If <math style="vertical-align: -19px">\lim_{x\rightarrow \infty} \frac{f'(x)}{g'(x)}</math>&thinsp; is finite or&thinsp; <math style="vertical-align: -4px">\pm \infty ,</math>
+&nbsp; &nbsp; &nbsp; &nbsp; If <math style="vertical-align: -19px">\lim_{x\rightarrow \infty} \frac{f'(x)}{g'(x)}</math>&thinsp; is finite or&thinsp; <math style="vertical-align: -4px">\pm \infty ,</math>
 |-
+|
-::then <math style="vertical-align: -19px">\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 style="vertical-align: -19px">\lim_{x\rightarrow \infty} \frac{f(x)}{g(x)}\,=\,\lim_{x\rightarrow \infty} \frac{f'(x)}{g'(x)}.</math>
 |}
@@ Line 34: / Line 32: @@
 |-
+|
-::<math>\lim_{x\rightarrow -3} \frac{x^3-9x}{6+2x}\,=\,\lim_{x\rightarrow -3}\frac{x(x-3)(x+3)}{2(x+3)}.</math>
+&nbsp; &nbsp; &nbsp; &nbsp;<math>\lim_{x\rightarrow -3} \frac{x^3-9x}{6+2x}\,=\,\lim_{x\rightarrow -3}\frac{x(x-3)(x+3)}{2(x+3)}.</math>
 |-
 |So, we can cancel <math style="vertical-align: -2px">x+3</math>&thinsp; in the numerator and denominator. Thus, we have
 |-
+|
-::<math>\lim_{x\rightarrow -3} \frac{x^3-9x}{6+2x}\,=\,\lim_{x\rightarrow -3}\frac{x(x-3)}{2}.</math>
+&nbsp; &nbsp; &nbsp; &nbsp;<math>\lim_{x\rightarrow -3} \frac{x^3-9x}{6+2x}\,=\,\lim_{x\rightarrow -3}\frac{x(x-3)}{2}.</math>
 |}
@@ Line 48: / Line 46: @@
 |-
+|
-::<math>\lim_{x\rightarrow -3} \frac{x^3-9x}{6+2x}\,=\,\frac{(-3)(-3-3)}{2}\,=\,\frac{18}{2}\,=\,9.</math>
+&nbsp; &nbsp; &nbsp; &nbsp;<math>\lim_{x\rightarrow -3} \frac{x^3-9x}{6+2x}\,=\,\frac{(-3)(-3-3)}{2}\,=\,\frac{18}{2}\,=\,9.</math>
 |}
@@ Line 59: / Line 57: @@
 |-
+|
-::<math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 \displaystyle{\lim_{x\rightarrow 0^+} \frac{\sin (2x)}{x^2}} & = & \displaystyle{\lim_{x\rightarrow 0^+}\frac{2\cos(2x)}{2x}}\\
 &&\\
@@ Line 69: / Line 67: @@
 !Step 2: &nbsp;
 |-
-|This limit is&thinsp; <math>+\infty.</math>
+|This limit is &nbsp; <math>+\infty.</math>
 |}
@@ Line 80: / Line 78: @@
 |-
+|
-::<math>\lim_{x\rightarrow -\infty} \frac{3x}{\sqrt{4x^2+x+5}}=\lim_{x\rightarrow -\infty} \frac{3x}{\sqrt{x^2(4+\frac{1}{x}+\frac{5}{x^2}})}.</math>
+&nbsp; &nbsp; &nbsp; &nbsp;<math>\lim_{x\rightarrow -\infty} \frac{3x}{\sqrt{4x^2+x+5}}=\lim_{x\rightarrow -\infty} \frac{3x}{\sqrt{x^2(4+\frac{1}{x}+\frac{5}{x^2}})}.</math>
 |-
 |Since we are looking at the limit as <math style="vertical-align: 0px">x</math> goes to negative infinity, we have <math style="vertical-align: -2px">\sqrt{x^2}=-x.</math>
@@ Line 87: / Line 85: @@
 |-
+|
-::<math>\lim_{x\rightarrow -\infty} \frac{3x}{\sqrt{4x^2+x+5}}\,=\,\lim_{x\rightarrow -\infty} \frac{3x}{-x\sqrt{4+\frac{1}{x}+\frac{5}{x^2}}}.</math>
+&nbsp; &nbsp; &nbsp; &nbsp;<math>\lim_{x\rightarrow -\infty} \frac{3x}{\sqrt{4x^2+x+5}}\,=\,\lim_{x\rightarrow -\infty} \frac{3x}{-x\sqrt{4+\frac{1}{x}+\frac{5}{x^2}}}.</math>
 |}
@@ Line 96: / Line 94: @@
 |-
+|
-::<math>\lim_{x\rightarrow -\infty} \frac{3x}{\sqrt{4x^2+x+5}}\,=\,\lim_{x\rightarrow -\infty} \frac{-3}{\sqrt{4+\frac{1}{x}+\frac{5}{x^2}}}.</math>
+&nbsp; &nbsp; &nbsp; &nbsp;<math>\lim_{x\rightarrow -\infty} \frac{3x}{\sqrt{4x^2+x+5}}\,=\,\lim_{x\rightarrow -\infty} \frac{-3}{\sqrt{4+\frac{1}{x}+\frac{5}{x^2}}}.</math>
 |-
 |So, we have
 |-
+|
-::<math>\lim_{x\rightarrow -\infty} \frac{3x}{\sqrt{4x^2+x+5}}=\frac{-3}{\sqrt{4}}=\frac{-3}{2}.</math>
+&nbsp; &nbsp; &nbsp; &nbsp;<math>\lim_{x\rightarrow -\infty} \frac{3x}{\sqrt{4x^2+x+5}}=\frac{-3}{\sqrt{4}}=\frac{-3}{2}.</math>
 |}
@@ Line 108: / Line 106: @@
 !Final Answer: &nbsp;
 |-
-|'''(a)'''&thinsp; <math style="vertical-align: 0px">9</math>
+|&nbsp; &nbsp; '''(a)'''&nbsp; &nbsp; <math style="vertical-align: 0px">9</math>
 |-
-|'''(b)'''&thinsp; <math style="vertical-align: 0px">+\infty</math>
+|&nbsp; &nbsp; '''(b)'''&nbsp; &nbsp; <math style="vertical-align: 0px">+\infty</math>
 |-
-|'''(c)'''&thinsp; <math style="vertical-align: -15px">-\frac{3}{2}</math>
+|&nbsp; &nbsp; '''(c)'''&nbsp; &nbsp; <math style="vertical-align: -15px">-\frac{3}{2}</math>
 |}
 [[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]