009A Sample Final 1, Problem 1 - Revision history

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MathAdmin: Created page with "In each part, compute the limit. If the limit is infinite, be sure to specify positive or negative infinity. a)
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Created page with "In each part, compute the limit. If the limit is infinite, be sure to specify positive or negative infinity. a) <math style="vertical-al..."

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In each part, compute the limit. If the limit is infinite, be sure to specify positive or negative infinity.

a) <math style="vertical-align: -14px">\lim_{x\rightarrow -3} \frac{x^3-9x}{6+2x}</math>

b) <math style="vertical-align: -14px">\lim_{x\rightarrow 0^+} \frac{\sin (2x)}{x^2}</math>

c) <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;"
!Foundations:  
|-
|Recall:
|-
|'''L'Hôpital's Rule'''
|-
|Suppose that <math style="vertical-align: -11px">\lim_{x\rightarrow \infty} f(x)</math>  and <math style="vertical-align: -11px">\lim_{x\rightarrow \infty} g(x)</math>  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>  is finite or  <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>
|}

'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We begin by factoring the numerator. We have
|-
|
::<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>  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>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we can just plug in <math style="vertical-align: -1px">x=-3</math>  to get
|-
|
::<math>\lim_{x\rightarrow -3} \frac{x^3-9x}{6+2x}\,=\,\frac{(-3)(-3-3)}{2}\,=\,\frac{18}{2}\,=\,9.</math>
|}

'''(b)'''

{| 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 0^+} \frac{\sin (2x)}{x^2}} & = & \displaystyle{\lim_{x\rightarrow 0^+}\frac{2\cos(2x)}{2x}}\\
&&\\
& = & \displaystyle{\lim_{x\rightarrow 0^+}\frac{\cos(2x)}{x}.}\\
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|This limit is  <math>+\infty.</math>
|}

'''(c)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
| We have
|-
|
::<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>
|-
|So, we have
|-
|
::<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>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
| We simplify to get
|-
|
::<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>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|'''(a)'''  <math style="vertical-align: 0px">9</math>
|-
|'''(b)'''  <math style="vertical-align: 0px">+\infty</math>
|-
|'''(c)'''  <math style="vertical-align: -15px">-\frac{3}{2}</math>
|}
[[009A_Sample_Final_1|'''Return to Sample Exam''']]

@@ 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 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) <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) <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) <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 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;"
 !Foundations: &nbsp;
 |-
-|Recall:
+|'''L'Hôpital's Rule'''
 |-
+|&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>
 |-
+|
-::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>
 |-
+|
-::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>
 |}
 '''Solution:'''
@@ Line 35: / 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
+|So, we can cancel &nbsp;<math style="vertical-align: -2px">x+3</math>&nbsp; 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 46: / 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
 |-
+|
-::<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 60: / Line 63: @@
 |-
+|
-::<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 70: / Line 73: @@
 !Step 2: &nbsp;
 |-
-|This limit is&thinsp; <math>+\infty.</math>
+|This limit is &nbsp; <math>\infty.</math>
 |}
@@ Line 81: / Line 84: @@
 |-
+|
-::<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>
+|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
 |-
+|
-::<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 97: / Line 100: @@
 |-
+|
-::<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}}}\\
+&&\\
-::<math>\lim_{x\rightarrow -\infty} \frac{3x}{\sqrt{4x^2+x+5}}=\frac{-3}{\sqrt{4}}=\frac{-3}{2}.</math>
+& = & \displaystyle{-\frac{3}{2}.}
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Final Answer: &nbsp;
 |-
-|&nbsp;&nbsp; '''(a)'''&thinsp; <math style="vertical-align: 0px">9</math>
+|&nbsp; &nbsp; '''(a)'''&nbsp; &nbsp; <math style="vertical-align: 0px">9</math>
 |-
-|&nbsp;&nbsp; '''(b)'''&thinsp; <math style="vertical-align: 0px">+\infty</math>
+|&nbsp; &nbsp; '''(b)'''&nbsp; &nbsp; <math style="vertical-align: 0px">\infty</math>
 |-
-|&nbsp;&nbsp; '''(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>''']]