Difference between revisions of "009A Sample Final 2, Problem 1"

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<span class="exam">(c) &nbsp;<math style="vertical-align: -15px">\lim_{x\rightarrow -\infty} \frac{\sqrt{x^2+2}}{2x-1}</math>
 
<span class="exam">(c) &nbsp;<math style="vertical-align: -15px">\lim_{x\rightarrow -\infty} \frac{\sqrt{x^2+2}}{2x-1}</math>
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[[009A Sample Final 2, Problem 1 Solution|'''<u>Solution</u>''']]
  
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!Foundations: &nbsp;
 
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|'''L'Hôpital's Rule'''
 
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|&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>
 
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&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>
 
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&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>
 
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[[009A Sample Final 2, Problem 1 Detailed Solution|'''<u>Detailed Solution</u>''']]
  
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'''Solution:'''
 
  
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'''(a)'''
 
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!Step 1: &nbsp;
 
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|We begin by noticing that we plug in &nbsp;<math style="vertical-align: 0px">x=4</math>&nbsp; into
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{\sqrt{x+5}-3}{x-4},</math>
 
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|we get &nbsp; <math style="vertical-align: -12px">\frac{0}{0}.</math>
 
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!Step 2: &nbsp;
 
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|Now, we multiply the numerator and denominator by the conjugate of the numerator.
 
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|Hence, we have
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 
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\displaystyle{\lim_{x\rightarrow 4} \frac{\sqrt{x+5}-3}{x-4}} & = & \displaystyle{\lim_{x\rightarrow 4} \frac{\sqrt{x+5}-3}{x-4}\frac{(\sqrt{x+5}+3)}{(\sqrt{x+5}+3)}}\\
 
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&&\\
 
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& = & \displaystyle{\lim_{x\rightarrow 4} \frac{(x+5)-9}{(x-4)(\sqrt{x+5}+3)}}\\
 
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&&\\
 
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& = & \displaystyle{\lim_{x\rightarrow 4} \frac{x-4}{(x-4)(\sqrt{x+5}+3)}}\\
 
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&&\\
 
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& = & \displaystyle{\lim_{x\rightarrow 4} \frac{1}{\sqrt{x+5}+3}}\\
 
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&&\\
 
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& = & \displaystyle{ \frac{1}{\sqrt{9}+3}}\\
 
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&&\\
 
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& = & \displaystyle{\frac{1}{6}.}
 
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\end{array}</math>
 
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'''(b)'''
 
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!Step 2: &nbsp;
 
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'''(c)'''
 
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!Step 1: &nbsp;
 
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!Step 2: &nbsp;
 
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!Final Answer: &nbsp;
 
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|&nbsp; &nbsp;'''(a)'''&nbsp; &nbsp;<math>\frac{1}{6}</math>
 
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|&nbsp; &nbsp;'''(b)'''&nbsp; &nbsp;<math>0</math>
 
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|&nbsp; &nbsp;'''(c)'''&nbsp; &nbsp;<math>\frac{-1}{2}</math>
 
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[[009A_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]
 
[[009A_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]

Latest revision as of 09:05, 1 December 2017

Compute

(a)  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \lim _{x\rightarrow 4}{\frac {{\sqrt {x+5}}-3}{x-4}}}

(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}{3x}}

(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{\sqrt{x^2+2}}{2x-1}}

Solution


Detailed Solution


Return to Sample Exam

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