009A Sample Final 3, Problem 1 - Revision history

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← Older revision		Revision as of 23:32, 2 December 2017
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	<span class="exam">(c)  <math style="vertical-align: -14px">\lim_{x\rightarrow -\infty} \frac{\sqrt{9x^6-x}}{3x^3+4x}</math>		<span class="exam">(c)  <math style="vertical-align: -14px">\lim_{x\rightarrow -\infty} \frac{\sqrt{9x^6-x}}{3x^3+4x}</math>

		+	<hr>
		+	[[009A Sample Final 3, Problem 1 Solution\|'''<u>Solution</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.'''  <math style="vertical-align: -14px">\lim_{x\rightarrow 0} \frac{\sin x}{x}=1</math>~~
−	\|}

		+	[[009A Sample Final 3, Problem 1 Detailed Solution\|'''<u>Detailed Solution</u>''']]

−	~~'''Solution:'''~~

−	~~'''(a)'''~~
−
−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Step 1:  ~~
−	\|-
−	~~\|We begin by noticing that we plug in  <math style="vertical-align: 0px">x=0</math>  into~~
−	\|-
−	~~\|        <math>\frac{\sin(5x)}{1-\sqrt{1-x}},</math>~~
−	\|-
−	~~\|we get   <math style="vertical-align: -12px">\frac{0}{0}.</math>~~
−	\|}
−
−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Step 2:  ~~
−	\|-
−	~~\|Now, we multiply the numerator and denominator by the conjugate of the denominator.~~
−	\|-
−	~~\|Hence, we have~~
−	\|-
−	~~\|        <math>\begin{array}{rcl}~~
−	~~\displaystyle{\lim_{x\rightarrow 0} \frac{\sin(5x)}{1-\sqrt{1-x}}} & = & \displaystyle{\lim_{x\rightarrow 0} \frac{\sin(5x)}{1-\sqrt{1-x}} \bigg(\frac{1+\sqrt{1-x}}{1+\sqrt{1-x}}\bigg)}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{\lim_{x\rightarrow 0} \frac{\sin(5x)(1+\sqrt{1-x})}{x}}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{\lim_{x\rightarrow 0} \frac{\sin(5x)}{x}(1+\sqrt{1-x})}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{\bigg(\lim_{x\rightarrow 0} \frac{\sin(5x)}{x}\bigg) \lim_{x\rightarrow 0}(1+\sqrt{1-x})}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{\bigg(5\lim_{x\rightarrow 0} \frac{\sin(5x)}{5x}\bigg) (2)}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{5(1)(2)}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{10.}~~
−	~~\end{array}</math>~~
−	\|}
−
−	~~'''(b)'''~~
−
−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Step 1:  ~~
−	\|-
−	~~\|Since  <math style="vertical-align: -12px">\lim_{x\rightarrow 8} 3 =3\ne 0,</math>~~
−	\|-
−	~~\|we have~~
−	\|-
−	~~\|        <math>\begin{array}{rcl}~~
−	~~\displaystyle{-2} & = & \displaystyle{\lim _{x\rightarrow 8} \bigg[\frac{xf(x)}{3}\bigg]}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{\frac{\displaystyle{\lim_{x\rightarrow 8} xf(x)}}{\displaystyle{\lim_{x\rightarrow 8} 3}}}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{\frac{\displaystyle{\lim_{x\rightarrow 8} xf(x)}}{3}.}~~
−	~~\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>3,</math>  we get~~
−	\|-
−	~~\|        <math>-6=\lim_{x\rightarrow 8} xf(x).</math>~~
−	\|-
−	~~\|Now, using properties of limits, we have~~
−	\|-
−	~~\|        <math>\begin{array}{rcl}~~
−	~~\displaystyle{-6} & = & \displaystyle{\bigg(\lim_{x\rightarrow 8} x\bigg)\bigg(\lim_{x\rightarrow 8}f(x)\bigg)}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{8\lim_{x\rightarrow 8} f(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 8} f(x)</math>  in the last equation,~~
−	\|-
−	~~\|we get~~
−	\|-
−	\|
−	~~        <math> \lim_{x\rightarrow 8} f(x)=-\frac{3}{4}.</math>~~
−	\|}
−
−	~~'''(c)'''~~
−
−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Step 1:  ~~
−	\|-
−	~~\|First, we write~~
−	\|-
−	~~\|        <math>\begin{array}{rcl}~~
−	\displaystyle{\lim_{x\rightarrow -\infty} \frac{\sqrt{9x^6-x}}{3x^3+4x}} & = & \displaystyle{\lim_{x\rightarrow -\infty} \frac{\sqrt{9x^6-x}}{3x^3+4x}\frac{\big(\frac{1}{x^3}\big)}{\big(\frac{1}{x^3}\big)}}\\
−	~~&&\\~~
−	~~& = & \displaystyle{\lim_{x\rightarrow -\infty} \frac{\sqrt{9-\frac{1}{x^5}}}{3+\frac{4}{x^2}}.}~~
−	~~\end{array}</math>~~
−	\|}
−
−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Step 2:  ~~
−	\|-
−	~~\|Now, we have~~
−	\|-
−	~~\|        <math>\begin{array}{rcl}~~
−	~~\displaystyle{\lim_{x\rightarrow -\infty} \frac{\sqrt{9x^6-x}}{3x^3+4x}} & = & \displaystyle{\frac{\lim_{x\rightarrow -\infty} \sqrt{9-\frac{1}{x^5}}}{\lim_{x\rightarrow -\infty}3+\frac{4}{x^2}}}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{\frac{\sqrt{9}}{3}}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{1.}~~
−	~~\end{array}</math>~~
−	\|}
−
−
−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Final Answer:  ~~
−	\|-
−	~~\|   '''(a)'''    <math>10</math>~~
−	\|-
−	~~\|   '''(b)'''    <math>-\frac{3}{4}</math>~~
−	\|-
−	~~\|   '''(c)'''    <math>1</math>~~
−	\|}
	[[009A_Sample_Final_3\|'''<u>Return to Sample Exam</u>''']]		[[009A_Sample_Final_3\|'''<u>Return to Sample Exam</u>''']]

@@ Line 69: / Line 69: @@
 \displaystyle{-2} & = & \displaystyle{\lim _{x\rightarrow 8} \bigg[\frac{xf(x)}{3}\bigg]}\\
 &&\\
-& = & \displaystyle{\frac{\lim_{x\rightarrow 8} xf(x)}{\lim_{x\rightarrow 8} 3}}\\
+& = & \displaystyle{\frac{\displaystyle{\lim_{x\rightarrow 8} xf(x)}}{\displaystyle{\lim_{x\rightarrow 8} 3}}}\\
 &&\\
-& = & \displaystyle{\frac{\lim_{x\rightarrow 8} xf(x)}{3}.}
+& = & \displaystyle{\frac{\displaystyle{\lim_{x\rightarrow 8} xf(x)}}{3}.}
 \end{array}</math>
 |-

@@ Line 82: / Line 82: @@
 |If we multiply both sides of the last equation by &nbsp;<math>3,</math>&nbsp; we get
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>-6=\lim_{x\rightarrow 8} xf(x)).</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>-6=\lim_{x\rightarrow 8} xf(x).</math>
 |-
 |Now, using properties of limits, we have
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-\displaystyle{10} & = & \displaystyle{\bigg(\lim_{x\rightarrow 8} x\bigg)\bigg(\lim_{x\rightarrow 8}f(x)\bigg)}\\
+\displaystyle{-6} & = & \displaystyle{\bigg(\lim_{x\rightarrow 8} x\bigg)\bigg(\lim_{x\rightarrow 8}f(x)\bigg)}\\
 &&\\
 & = & \displaystyle{8\lim_{x\rightarrow 8} f(x).}\\
@@ Line 101: / Line 101: @@
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp; <math> \lim_{x\rightarrow 8} f(x)=\frac{-3}{4}.</math>
+&nbsp; &nbsp; &nbsp; &nbsp; <math> \lim_{x\rightarrow 8} f(x)=-\frac{3}{4}.</math>
 |}
@@ Line 138: / Line 138: @@
 |&nbsp; &nbsp;'''(a)'''&nbsp; &nbsp; <math>10</math>
 |-
-|&nbsp; &nbsp;'''(b)'''&nbsp; &nbsp; <math>\frac{-3}{4}</math>
+|&nbsp; &nbsp;'''(b)'''&nbsp; &nbsp; <math>-\frac{3}{4}</math>
 |-
 |&nbsp; &nbsp;'''(c)'''&nbsp; &nbsp; <math>1</math>
 |}
 [[009A_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]

@@ Line 26: / Line 26: @@
 !Step 1: &nbsp;
 |-
+|We begin by noticing that we plug in &nbsp;<math style="vertical-align: 0px">x=0</math>&nbsp; into
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{\sin(5x)}{1-\sqrt{1-x}},</math>
 |-
+|we get &nbsp; <math style="vertical-align: -12px">\frac{0}{0}.</math>
 |}
@@ Line 38: / Line 36: @@
 !Step 2: &nbsp;
 |-
+|Now, we multiply the numerator and denominator by the conjugate of the denominator.
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 |}
@@ Line 122: / Line 136: @@
 !Final Answer: &nbsp;
 |-
-|'''(a)'''
+|&nbsp; &nbsp;'''(a)'''&nbsp; &nbsp; <math>10</math>
 |-
 |&nbsp; &nbsp;'''(b)'''&nbsp; &nbsp; <math>\frac{-3}{4}</math>

@@ Line 95: / Line 95: @@
 !Step 1: &nbsp;
 |-
+|First, we write
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-|-
+\displaystyle{\lim_{x\rightarrow -\infty} \frac{\sqrt{9x^6-x}}{3x^3+4x}} & = & \displaystyle{\lim_{x\rightarrow -\infty} \frac{\sqrt{9x^6-x}}{3x^3+4x}\frac{\big(\frac{1}{x^3}\big)}{\big(\frac{1}{x^3}\big)}}\\
+&&\\
-|-
+& = & \displaystyle{\lim_{x\rightarrow -\infty} \frac{\sqrt{9-\frac{1}{x^5}}}{3+\frac{4}{x^2}}.}
+\end{array}</math>
 |}
@@ Line 109: / Line 107: @@
 !Step 2: &nbsp;
 |-
+|Now, we have
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 |}
@@ Line 126: / Line 126: @@
 |&nbsp; &nbsp;'''(b)'''&nbsp; &nbsp; <math>\frac{-3}{4}</math>
 |-
-|'''(c)'''
+|&nbsp; &nbsp;'''(c)'''&nbsp; &nbsp; <math>1</math>
 |}
 [[009A_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]

@@ Line 11: / Line 11: @@
 !Foundations: &nbsp;
 |-
+| '''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.''' &nbsp;<math style="vertical-align: -14px">\lim_{x\rightarrow 0} \frac{\sin x}{x}=1</math>
 |}

@@ Line 1: / Line 1: @@
 <span class="exam">Find each of the following limits if it exists. If you think the limit does not exist provide a reason.
-<span class="exam">(a) <math style="vertical-align: -14px">\lim_{x\rightarrow 0} \frac{\sin(5x)}{1-\sqrt{1-x}}</math>
+<span class="exam">(a) &nbsp;<math style="vertical-align: -14px">\lim_{x\rightarrow 0} \frac{\sin(5x)}{1-\sqrt{1-x}}</math>
-<span class="exam">(b) <math style="vertical-align: -14px">\lim_{x\rightarrow 8} f(x),</math> given that <math>\lim_{x\rightarrow 8}\frac{xf(x)}{3}=-2</math>
+<span class="exam">(b) &nbsp;<math style="vertical-align: -12px">\lim_{x\rightarrow 8} f(x),</math>&nbsp; given that &nbsp;<math style="vertical-align: -14px">\lim_{x\rightarrow 8}\frac{xf(x)}{3}=-2</math>
-<span class="exam">(c) <math style="vertical-align: -14px">\lim_{x\rightarrow -\infty} \frac{\sqrt{9x^6-x}}{3x^3+4x}</math>
+<span class="exam">(c) &nbsp;<math style="vertical-align: -14px">\lim_{x\rightarrow -\infty} \frac{\sqrt{9x^6-x}}{3x^3+4x}</math>

@@ Line 1: / Line 1: @@
 <span class="exam">Find each of the following limits if it exists. If you think the limit does not exist provide a reason.
-::<span class="exam">a) <math style="vertical-align: -14px">\lim_{x\rightarrow 0} \frac{\sin(5x)}{1-\sqrt{1-x}}</math>
+<span class="exam">(a) <math style="vertical-align: -14px">\lim_{x\rightarrow 0} \frac{\sin(5x)}{1-\sqrt{1-x}}</math>
-::<span class="exam">b) <math style="vertical-align: -14px">\lim_{x\rightarrow 8} f(x),</math> given that <math>\lim_{x\rightarrow 8}\frac{xf(x)}{3}=-2</math>
+<span class="exam">(b) <math style="vertical-align: -14px">\lim_{x\rightarrow 8} f(x),</math> given that <math>\lim_{x\rightarrow 8}\frac{xf(x)}{3}=-2</math>
-::<span class="exam">c) <math style="vertical-align: -14px">\lim_{x\rightarrow -\infty} \frac{\sqrt{9x^6-x}}{3x^3+4x}</math>
+<span class="exam">(c) <math style="vertical-align: -14px">\lim_{x\rightarrow -\infty} \frac{\sqrt{9x^6-x}}{3x^3+4x}</math>
@@ Line 19: / Line 19: @@
+|
 |}
 '''Solution:'''
@@ Line 87: / Line 88: @@
+|
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"