009A Sample Final A, Problem 1 - Revision history

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MathAdmin: Created page with " 1. Find the following limits:
(a) $\lim_{x\rightarrow0}\frac{\ta..."$

2015-03-27T17:12:20Z

Created page with " 1. Find the following limits: (a) <math style="vertical-align: -45%;">\lim_{x\rightarrow0}\frac{\ta..."

New page

1. Find the following limits:    (a)   <math style="vertical-align: -45%;">\lim_{x\rightarrow0}\frac{\tan(3x)}{x^{3}}.</math>
 
   (b) <math style="vertical-align: -52%;">\lim_{x\rightarrow-\infty}\frac{\sqrt{x^{6}+6x^{2}+2}}{x^{3}+x-1}.</math>
 
   (c)   <math style="vertical-align: -65%;">\lim_{x\rightarrow3}\frac{x-3}{\sqrt{x+1}-2}.</math>
 
   (d)   <math style="vertical-align: -65%;">\lim_{x\rightarrow3}\frac{x-1}{\sqrt{x+1}-1}.</math>
 
   (e)  <math style="vertical-align: -50%;">\lim_{x\rightarrow\infty}\frac{5x^{2}-2x+3}{1-3x^{2}}.</math>
 

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|When evaluating limits of rational functions, the first idea to try is to simply plug in the limit. Unfortunately, most (but not all) exam questions require more work. Many of them will evaluate to an '''indeterminate form''', or something of the form
|-
|      <math>\frac{0}{0}</math>   or   <math>\frac {\pm \infty}{\pm \infty}.</math>
|-
| In this case, here are several approaches to try:
|-
|
*We can multiply the numerator and denominator by the conjugate of the denominator. This frequently results in a term that cancels, allowing us to then just plug in our limit value.
*We can factor a term creatively. For example, <math style="vertical-align: -5%;">x-1</math> can be factored as <math style="vertical-align: -45%;">\left(\sqrt{x}-1\right) \left(\sqrt{x}+1\right)</math> , or as <math style="vertical-align: -60%;">\left(\sqrt[3]{x}-1\right)\left(\left(\sqrt[3]{x}\right)^{2}+\sqrt[3]{x}+1\right)</math> , both of which could result in a factor that cancels in our fraction.
*We can apply '''l'Hôpital's Rule:''' Suppose <math style="vertical-align: 0%;">c</math> is contained in some interval <math style="vertical-align: 0%;">I</math>. If <math style="vertical-align: -50%;">\lim_{x \to c}f(x)=\lim_{x \to c}g(x)=0 \text{ or } \pm\infty</math>  and <math style="vertical-align: -75%;">\lim_{x\to c}\frac{f'(x)}{g'(x)}</math>   exists, and <math style="vertical-align: -20%;">g'(x)\neq 0</math>  for all <math style="vertical-align: -20%;">x\neq c</math> in <math style="vertical-align: 0%;">I</math>, then <math style="vertical-align: -73%;">\lim_{x\to c}\frac{f(x)}{g(x)} = \lim_{x\to c}\frac{f'(x)}{g'(x)}</math>.
|-
|Note that the first requirement in l'Hôpital's Rule is that the fraction ''must'' be an indeterminate form. This should be shown in your answer for any exam question. 
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Part (a):  

|}

[[009A_Sample_Final_A|'''Return to Sample Exam''']]

@@ Line 36: / Line 36: @@
 |}
-'''Solution:'''
+&nbsp;'''Solution:'''
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"

@@ Line 35: / Line 35: @@
 |Note that the first requirement in l'H&ocirc;pital's Rule is that the fraction <u>''must''</u> be an indeterminate form.  This should be shown in your answer for any exam question.<br>
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"

@@ Line 26: / Line 26: @@
 |<br>&nbsp;&nbsp;&nbsp;&nbsp; <math>\frac{0}{0}</math>&nbsp;&nbsp; or &nbsp;&nbsp;<math>\frac {\pm \infty}{\pm \infty}.</math>
 |-
-|<br>In this case, here are several approaches to try:
+|<br>In this case, there are several approaches to try:
 |-
+|

@@ Line 87: / Line 87: @@
 |Alternatively, we can apply l'H&ocirc;pital's Rule:
 |-
-|&nbsp;&nbsp;&nbsp;&nbsp; <math>\lim_{x\rightarrow3}\frac{x-3}{\sqrt{x+1}-2}\,\,\overset{l'H}{=}\,\,\lim_{x\rightarrow3}\frac{1}{{\displaystyle \frac{1}{2}\cdot\frac{1}{\sqrt{x+1}}}}\,=\,\frac{1}{\frac{1}{2}\cdot\frac{1}{2}}\,=\,4.</math>
+|&nbsp;&nbsp;&nbsp;&nbsp; <math>\lim_{x\rightarrow3}\frac{x-3}{\sqrt{x+1}-2}\,\,\overset{l'H}{=}\,\,\lim_{x\rightarrow3}\frac{1}{{\displaystyle \frac{1}{2}\cdot\frac{1}{\sqrt{x+1}}}}\,=\,\frac{1}{\frac{1}{2}\cdot\frac{1}{2}}\,=\,4.</math><br>
+|}
 |}
 [[009A_Sample_Final_A|'''<u>Return to Sample Exam</u>''']]

@@ Line 69: / Line 69: @@
 |&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <math>=\,-1.</math><br>
 |}
 [[009A_Sample_Final_A|'''<u>Return to Sample Exam</u>''']]

@@ Line 14: / Line 14: @@
 !Foundations: &nbsp;
 |-
-|When evaluating limits of rational functions, the first idea to try is to simply plug in the limit.  Unfortunately, most (but not all) exam questions require more work.  Many of them will evaluate to an '''indeterminate form''', or something of the form
+|When evaluating limits of rational functions, the first idea to try is to simply plug in the limit. In addition to this, we must consider that as a limit,
 |-
 |<br>&nbsp;&nbsp;&nbsp;&nbsp; <math>\frac{0}{0}</math>&nbsp;&nbsp; or &nbsp;&nbsp;<math>\frac {\pm \infty}{\pm \infty}.</math>
@@ Line 30: / Line 38: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Part (a): &nbsp;
 |}
 [[009A_Sample_Final_A|'''<u>Return to Sample Exam</u>''']]

@@ Line 22: / Line 22: @@
+|
 *We can multiply the numerator and denominator by the conjugate of the denominator.  This frequently results in a term that cancels, allowing us to then just plug in our limit value.
-*We can factor a term creatively.  For example, <math style="vertical-align: -8%;">x-1</math> can be factored as <math style="vertical-align: -48%;">\left(\sqrt{x}-1\right) \left(\sqrt{x}+1\right)</math>&thinsp;, or as <math style="vertical-align: -70%;">\left(\sqrt[3]{x}-1\right)\left(\left(\sqrt[3]{x}\right)^{2}+\sqrt[3]{x}+1\right)</math>&thinsp;, both of which could result in a factor that cancels in our fraction.
+*We can factor a term creatively.  For example, <math style="vertical-align: -9%;">x-1</math> can be factored as <math style="vertical-align: -48%;">\left(\sqrt{x}-1\right) \left(\sqrt{x}+1\right)</math>&thinsp;, or as <math style="vertical-align: -71%;">\left(\sqrt[3]{x}-1\right)\left(\left(\sqrt[3]{x}\right)^{2}+\sqrt[3]{x}+1\right)</math>&thinsp;, both of which could result in a factor that cancels in our fraction.
-*We can apply '''l'H&ocirc;pital's Rule:''' Suppose <math style="vertical-align: 0%;">c</math> is contained in some interval <math style="vertical-align: 0%;">I</math>. If <math style="vertical-align: -60%;">\lim_{x \to c}f(x)=\lim_{x \to c}g(x)=0 \text{ or } \pm\infty</math>&thinsp; and <math style="vertical-align: -85%;">\lim_{x\to c}\frac{f'(x)}{g'(x)}</math> &nbsp; exists, and <math style="vertical-align: -25%;">g'(x)\neq 0</math>&thinsp; for all <math style="vertical-align: -23%;">x\neq c</math> in <math style="vertical-align: 0%;">I</math>, then <math style="vertical-align: -83%;">\lim_{x\to c}\frac{f(x)}{g(x)} = \lim_{x\to c}\frac{f'(x)}{g'(x)}</math>.
+*We can apply '''l'H&ocirc;pital's Rule:''' Suppose <math style="vertical-align: 0%;">c</math> is contained in some interval <math style="vertical-align: 0%;">I</math>. If <math style="vertical-align: -60%;">\lim_{x \to c}f(x)=\lim_{x \to c}g(x)=0 \text{ or } \pm\infty</math>&thinsp; and <math style="vertical-align: -85%;">\lim_{x\to c}\frac{f'(x)}{g'(x)}</math> &nbsp; exists, and <math style="vertical-align: -24%;">g'(x)\neq 0</math>&thinsp; for all <math style="vertical-align: -23%;">x\neq c</math>&thinsp; in <math style="vertical-align: 0%;">I</math>, then <math style="vertical-align: -84%;">\lim_{x\to c}\frac{f(x)}{g(x)} = \lim_{x\to c}\frac{f'(x)}{g'(x)}</math>.
 |-
 |Note that the first requirement in l'H&ocirc;pital's Rule is that the fraction <u>''must''</u> be an indeterminate form.  This should be shown in your answer for any exam question.<br>

009A Sample Final A, Problem 1 - Revision history

MathAdmin at 02:21, 28 March 2015

MathAdmin at 22:22, 27 March 2015

MathAdmin at 22:22, 27 March 2015

MathAdmin at 19:05, 27 March 2015

MathAdmin at 18:57, 27 March 2015

MathAdmin at 18:28, 27 March 2015

MathAdmin at 17:18, 27 March 2015

MathAdmin at 17:15, 27 March 2015

MathAdmin: Created page with " 1. Find the following limits: (a) \lim_{x\rightarrow0}\frac{\ta..."

MathAdmin: Created page with " 1. Find the following limits:
(a) $\lim_{x\rightarrow0}\frac{\ta..."$