Difference between revisions of "009A Sample Final 1, Problem 1"

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(Created page with "<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-al...")
 
 
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<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">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;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Foundations: &nbsp;  
 
!Foundations: &nbsp;  
 
|-
 
|-
|Recall:
+
|'''L'Hôpital's Rule, Part 1'''
 
|-
 
|-
|'''L'Hôpital's Rule'''
+
|
|-
+
&nbsp; &nbsp; &nbsp; &nbsp; Let &nbsp;<math style="vertical-align: -12px">\lim_{x\rightarrow c}f(x)=0</math>&nbsp; and &nbsp;<math style="vertical-align: -12px">\lim_{x\rightarrow c}g(x)=0,</math>&nbsp; where &nbsp;<math style="vertical-align: -5px">f</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">g</math>&nbsp; are differentiable functions
|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;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;
::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;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>
::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:'''
 
'''Solution:'''
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|-
 
|-
 
|
 
|
::<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>
 
|}
 
|}
  
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!Step 2: &nbsp;
 
!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}
 +
\displaystyle{\lim_{x\rightarrow -3} \frac{x^3-9x}{6+2x}} & = & \displaystyle{\frac{(-3)(-3-3)}{2}}\\
 +
&&\\
 +
& = & \displaystyle{\frac{18}{2}}\\
 +
&&\\
 +
& = & \displaystyle{9.}
 +
\end{array}</math>
 
|}
 
|}
  
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|-
 
|-
 
|
 
|
::<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}}\\
 
\displaystyle{\lim_{x\rightarrow 0^+} \frac{\sin (2x)}{x^2}} & = & \displaystyle{\lim_{x\rightarrow 0^+}\frac{2\cos(2x)}{2x}}\\
 
&&\\
 
&&\\
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
|This limit is&thinsp; <math>+\infty.</math>
+
|This limit is &nbsp; <math>\infty.</math>
 
|}
 
|}
  
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|-
 
|-
 
|
 
|
::<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
 
|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>
 
|}
 
|}
  
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|-
 
|-
 
|
 
|
::<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}.}
 +
\end{array}</math>
 
|}
 
|}
 +
  
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Final Answer: &nbsp;  
 
!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>''']]
 
[[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

Latest revision as of 18:10, 20 May 2017

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

(a)  

(b)  

(c)  

Foundations:  
L'Hôpital's Rule, Part 1

        Let    and    where    and    are differentiable functions

       on an open interval    containing    and    on    except possibly at   
       Then,  


Solution:

(a)

Step 1:  
We begin by factoring the numerator. We have

       

So, we can cancel    in the numerator and denominator. Thus, we have

       

Step 2:  
Now, we can just plug in    to get

       

(b)

Step 1:  
We proceed using L'Hôpital's Rule. So, we have

       

Step 2:  
This limit is  

(c)

Step 1:  
We have

       

Since we are looking at the limit as    goes to negative infinity, we have  
So, we have

       

Step 2:  
We simplify to get

       


Final Answer:  
    (a)   
    (b)   
    (c)   

Return to Sample Exam