009A Sample Final 2, Problem 8 - Revision history

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← Older revision		Revision as of 17:23, 1 December 2017
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	<span class="exam">(c)  <math style="vertical-align: -15px">\lim_{x\rightarrow 1} \frac{x^3-1}{x^{10}-1}</math>		<span class="exam">(c)  <math style="vertical-align: -15px">\lim_{x\rightarrow 1} \frac{x^3-1}{x^{10}-1}</math>
		+	<hr>
		+	[[009A Sample Final 2, Problem 8 Solution\|'''<u>Solution</u>''']]

−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Foundations:  ~~
−	\|-
−	~~\|'''L'Hôpital's Rule, Part 1'''~~
−	\|-
−	\|
−	Let  <math style="vertical-align: -12px">\lim_{x\rightarrow c}f(x)=0</math>  and  <math style="vertical-align: -12px">\lim_{x\rightarrow c}g(x)=0,</math>  where  <math style="vertical-align: -5px">f</math>  and  <math style="vertical-align: -5px">g</math>  are differentiable functions
−	\|-
−	\|       on an open interval  <math style="vertical-align: 0px">I</math>  containing  <math style="vertical-align: -5px">c,</math>  and  <math style="vertical-align: -5px">g'(x)\ne 0</math>  on  <math style="vertical-align: 0px">I</math>  except possibly at  <math style="vertical-align: 0px">c.</math>
−	\|-
−	~~\|       Then,   <math style="vertical-align: -18px">\lim_{x\rightarrow c} \frac{f(x)}{g(x)}=\lim_{x\rightarrow c} \frac{f'(x)}{g'(x)}.</math>~~
−	\|}

		+	[[009A Sample Final 2, Problem 8 Detailed Solution\|'''<u>Detailed Solution</u>''']]

−	~~'''Solution:'''~~

−	~~'''(a)'''~~
−
−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Step 1:  ~~
−	\|-
−	~~\|First, we have~~
−	\|-
−	~~\|        <math>\begin{array}{rcl}~~
−	~~\displaystyle{\lim_{x\rightarrow \infty} \frac{x^{-1}+x}{1+\sqrt{1+x}}} & = & \displaystyle{\lim_{x\rightarrow \infty}\frac{\frac{1}{x}+x}{1+\sqrt{1+x}}}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{\lim_{x\rightarrow \infty}\frac{\frac{1}{x}+x}{1+\sqrt{1+x}} \frac{\big(\frac{1}{\sqrt{x}}\big)}{\big(\frac{1}{\sqrt{x}}\big)}}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{\lim_{x\rightarrow \infty} \frac{\frac{1}{x^{3/2}}+\sqrt{x}}{\frac{1}{\sqrt{x}}+\sqrt{\frac{1}{x}+1}}.}~~
−	~~\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{x^{-1}+x}{1+\sqrt{1+x}}} & = & \displaystyle{\lim_{x\rightarrow \infty} \frac{\frac{1}{x^{3/2}}+\sqrt{x}}{\frac{1}{\sqrt{x}}+\sqrt{\frac{1}{x}+1}}}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{\frac{\lim_{x\rightarrow \infty}\big(\frac{1}{x^{3/2}}+\sqrt{x}\big)}{\lim_{x\rightarrow \infty}\big(\frac{1}{\sqrt{x}}+\sqrt{\frac{1}{x}+1}\big)}}\\~~
−	~~&&\\~~
−	& = & \displaystyle{\frac{\lim_{x\rightarrow \infty}\frac{1}{x^{3/2}}+\lim_{x\rightarrow \infty}\sqrt{x}}{\lim_{x\rightarrow \infty}\frac{1}{\sqrt{x}}+\lim_{x\rightarrow \infty}\sqrt{\frac{1}{x}+1}}}\\
−	~~&&\\~~
−	~~& = & \displaystyle{\frac{0+\lim_{x\rightarrow \infty}\sqrt{x}}{0+1}}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{\infty.}~~
−	~~\end{array}</math>~~
−	\|}
−
−	~~'''(b)'''~~
−
−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Step 1:  ~~
−	\|-
−	~~\|First, we write~~
−	\|-
−	~~\|        <math>\begin{array}{rcl}~~
−	~~\displaystyle{\lim_{x\rightarrow 0} \frac{\sin x}{\cos x-1}} & = & \displaystyle{\lim_{x\rightarrow 0} \frac{\sin x}{\cos x-1}\frac{(\cos x+1)}{(\cos x+1)}}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{\lim_{x\rightarrow 0} \frac{\sin x (\cos x+1)}{\cos^2x-1}}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{\lim_{x\rightarrow 0} \frac{\sin x(\cos x+1)}{-\sin^2 x}}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{\lim_{x\rightarrow 0} \frac{\cos x+1}{-\sin x}.}~~
−	~~\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 0^+} \frac{\sin x}{\cos x-1}} & = & \displaystyle{\lim_{x\rightarrow 0^+} \frac{\cos x+1}{-\sin x}}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{-\infty}~~
−	~~\end{array}</math>~~
−	\|-
−	~~\|and~~
−	\|-
−	~~\|        <math>\begin{array}{rcl}~~
−	~~\displaystyle{\lim_{x\rightarrow 0^-} \frac{\sin x}{\cos x-1}} & = & \displaystyle{\lim_{x\rightarrow 0^-} \frac{\cos x+1}{-\sin x}}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{\infty.}~~
−	~~\end{array}</math>~~
−	\|-
−	~~\|Therefore,~~
−	\|-
−	~~\|       <math>\lim_{x\rightarrow 0} \frac{\sin x}{\cos x-1}=\text{DNE}.</math>~~
−	\|}
−
−	~~'''(c)'''~~
−
−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Step 1:  ~~
−	\|-
−	~~\|We proceed using L'Hôpital's Rule. So, we have~~
−	\|-
−	\|
−	~~       <math>\begin{array}{rcl}~~
−	~~\displaystyle{\lim_{x\rightarrow 1} \frac{x^3-1}{x^{10}-1}} & \overset{L'H}{=} & \displaystyle{\lim_{x\rightarrow 1}\frac{3x^2}{10x^9}.}~~
−	~~\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 1} \frac{x^3-1}{x^{10}-1}} & \overset{L'H}{=} & \displaystyle{\lim_{x\rightarrow 1}\frac{3x^2}{10x^9}}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{\frac{3(1)^2}{10(1)^9}}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{\frac{3}{10}.}~~
−	~~\end{array}</math>~~
−	\|}
−
−
−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Final Answer:  ~~
−	\|-
−	~~\|   '''(a)'''    <math>\infty</math>~~
−	\|-
−	~~\|   '''(b)'''    <math>\text{DNE}</math>~~
−	\|-
−	~~\|   '''(c)'''    <math>\frac{3}{10}</math>~~
−	\|}
	[[009A_Sample_Final_2\|'''<u>Return to Sample Exam</u>''']]		[[009A_Sample_Final_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 29: / Line 29: @@
 !Step 1: &nbsp;
 |-
+|First, we have
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-|-
+\displaystyle{\lim_{x\rightarrow \infty} \frac{x^{-1}+x}{1+\sqrt{1+x}}} & = & \displaystyle{\lim_{x\rightarrow \infty}\frac{\frac{1}{x}+x}{1+\sqrt{1+x}}}\\
+&&\\
-|-
+& = & \displaystyle{\lim_{x\rightarrow \infty}\frac{\frac{1}{x}+x}{1+\sqrt{1+x}} \frac{\big(\frac{1}{\sqrt{x}}\big)}{\big(\frac{1}{\sqrt{x}}\big)}}\\
+&&\\
 |}
@@ Line 41: / Line 43: @@
 !Step 2: &nbsp;
 |-
+|Now, we have
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 |}
@@ Line 119: / Line 131: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp; &nbsp;'''(a)'''&nbsp; &nbsp;
+|&nbsp; &nbsp;'''(a)'''&nbsp; &nbsp; <math>\infty</math>
 |-
 |&nbsp; &nbsp;'''(b)'''&nbsp; &nbsp; <math>\text{DNE}</math>

@@ Line 10: / Line 10: @@
 !Foundations: &nbsp;
 |-
-|<span class="exam">In each part, compute the limit. If the limit is infinite, be sure to specify positive or negative infinity.
+|'''L'Hôpital's Rule'''
+|-
-<span class="exam">(a) &nbsp; <math style="vertical-align: -14px">\lim_{x\rightarrow -3} \frac{x^3-9x}{6+2x}</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>
+|-
-<span class="exam">(b) &nbsp; <math style="vertical-align: -14px">\lim_{x\rightarrow 0^+} \frac{\sin (2x)}{x^2}</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>
-<span class="exam">(c) &nbsp; <math style="vertical-align: -14px">\lim_{x\rightarrow -\infty} \frac{3x}{\sqrt{4x^2+x+5}}</math>
+|-
 |}

@@ Line 49: / Line 49: @@
 !Step 1: &nbsp;
 |-
-|We begin by noticing that we plug in &nbsp;<math style="vertical-align: 0px">x=0</math>&nbsp; into
+|First, we write
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{\sin x}{\cos x-1},</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-|-
+\displaystyle{\lim_{x\rightarrow 0} \frac{\sin x}{\cos x-1}} & = & \displaystyle{\lim_{x\rightarrow 0} \frac{\sin x}{\cos x-1}\frac{(\cos x+1)}{(\cos x+1)}}\\
-|we get &nbsp; <math style="vertical-align: -12px">\frac{0}{0}.</math>
+&&\\
-|-
+& = & \displaystyle{\lim_{x\rightarrow 0} \frac{\sin x (\cos x+1)}{\cos^2x-1}}\\
+&&\\
 |}
@@ Line 61: / Line 65: @@
 !Step 2: &nbsp;
 |-
+|Now, we have
 |}
@@ Line 95: / Line 117: @@
 !Final Answer: &nbsp;
 |-
-|'''(a)'''
+|&nbsp; &nbsp;'''(a)'''&nbsp; &nbsp;
 |-
-|'''(b)'''
+|&nbsp; &nbsp;'''(b)'''&nbsp; &nbsp; <math>\text{DNE}</math>
 |-
 |&nbsp; &nbsp;'''(c)'''&nbsp; &nbsp; <math>\frac{3}{10}</math>
 |}
 [[009A_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 1: / Line 1: @@
 <span class="exam">Compute
-<span class="exam">(a) <math>\lim_{x\rightarrow \infty} \frac{x^{-1}+x}{1+\sqrt{1+x}}</math>
+<span class="exam">(a) &nbsp;<math style="vertical-align: -18px">\lim_{x\rightarrow \infty} \frac{x^{-1}+x}{1+\sqrt{1+x}}</math>
-<span class="exam">(b) <math>\lim_{x\rightarrow 0} \frac{\sin x}{\cos x-1}</math>
+<span class="exam">(b) &nbsp;<math style="vertical-align: -15px">\lim_{x\rightarrow 0} \frac{\sin x}{\cos x-1}</math>
-<span class="exam">(c) <math>\lim_{x\rightarrow 1} \frac{x^3-1}{x^{10}-1}</math>
+<span class="exam">(c) &nbsp;<math style="vertical-align: -15px">\lim_{x\rightarrow 1} \frac{x^3-1}{x^{10}-1}</math>
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"