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

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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Foundations:    
 
!Foundations:    
−
|-
 
−
|Recall:
 
 
|-
 
|-
 
|'''L'Hôpital's Rule'''  
 
|'''L'Hôpital's Rule'''  
 
|-
 
|-
−
|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; 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>
 
|-
 
|-
 
|
 
|
−
::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 <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>
 
|-
 
|-
 
|
 
|
−
::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 <math style="vertical-align: -19px">\lim_{x\rightarrow \infty} \frac{f(x)}{g(x)}\,=\,\lim_{x\rightarrow \infty} \frac{f'(x)}{g'(x)}.</math>
 
|}
 
|}
  
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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 <math style="vertical-align: -2px">x+3</math>&thinsp; 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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|-
 
|-
 
|
 
|
−
::<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>\lim_{x\rightarrow -3} \frac{x^3-9x}{6+2x}\,=\,\frac{(-3)(-3-3)}{2}\,=\,\frac{18}{2}\,=\,9.</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 <math style="vertical-align: 0px">x</math> goes to negative infinity, we have <math style="vertical-align: -2px">\sqrt{x^2}=-x.</math>
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|-
 
|-
 
|
 
|
−
::<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>\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
 
|So, we have
 
|-
 
|-
 
|
 
|
−
::<math>\lim_{x\rightarrow -\infty} \frac{3x}{\sqrt{4x^2+x+5}}=\frac{-3}{\sqrt{4}}=\frac{-3}{2}.</math>
+
&nbsp; &nbsp; &nbsp; &nbsp;<math>\lim_{x\rightarrow -\infty} \frac{3x}{\sqrt{4x^2+x+5}}=\frac{-3}{\sqrt{4}}=\frac{-3}{2}.</math>
 
|}
 
|}
  
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!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>''']]

Revision as of 16:25, 25 February 2017

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

(a) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\rightarrow -3} \frac{x^3-9x}{6+2x}}

(b) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\rightarrow 0^+} \frac{\sin (2x)}{x^2}}

(c) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\rightarrow -\infty} \frac{3x}{\sqrt{4x^2+x+5}}}

Foundations:  
L'Hôpital's Rule
        Suppose that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\rightarrow \infty} f(x)}   and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\rightarrow \infty} g(x)}   are both zero or both Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \pm \infty .}

        If Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\rightarrow \infty} \frac{f'(x)}{g'(x)}}   is finite or  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \pm \infty ,}

        then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\rightarrow \infty} \frac{f(x)}{g(x)}\,=\,\lim_{x\rightarrow \infty} \frac{f'(x)}{g'(x)}.}


Solution:

(a)

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

       Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\rightarrow -3} \frac{x^3-9x}{6+2x}\,=\,\lim_{x\rightarrow -3}\frac{x(x-3)(x+3)}{2(x+3)}.}

So, we can cancel Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x+3}   in the numerator and denominator. Thus, we have

       Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\rightarrow -3} \frac{x^3-9x}{6+2x}\,=\,\lim_{x\rightarrow -3}\frac{x(x-3)}{2}.}

Step 2:  
Now, we can just plug in Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=-3}   to get

       Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\rightarrow -3} \frac{x^3-9x}{6+2x}\,=\,\frac{(-3)(-3-3)}{2}\,=\,\frac{18}{2}\,=\,9.}

(b)

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

       Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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}}

Step 2:  
This limit is   Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle +\infty.}

(c)

Step 1:  
We have

       Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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}})}.}

Since we are looking at the limit as Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} goes to negative infinity, we have Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sqrt{x^2}=-x.}
So, we have

       Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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}}}.}

Step 2:  
We simplify to get

       Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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}}}.}

So, we have

       Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\rightarrow -\infty} \frac{3x}{\sqrt{4x^2+x+5}}=\frac{-3}{\sqrt{4}}=\frac{-3}{2}.}


Final Answer:  
    (a)    Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 9}
    (b)    Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle +\infty}
    (c)    Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\frac{3}{2}}

Return to Sample Exam

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