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

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(Created page with "<span class="exam">Find the following limits: <span class="exam">(a) Find  <math style="vertical-align: -13px">\lim _{x\rightarrow 2} g(x),</math>  provided that &n...")
 
 
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<span class="exam">(c) Evaluate &nbsp;<math style="vertical-align: -14px">\lim _{x\rightarrow -3^+} \frac{x}{x^2-9} </math>
 
<span class="exam">(c) Evaluate &nbsp;<math style="vertical-align: -14px">\lim _{x\rightarrow -3^+} \frac{x}{x^2-9} </math>
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[[009A Sample Midterm 1, Problem 1 Solution|'''<u>Solution</u>''']]
  
  
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[[009A Sample Midterm 1, Problem 1 Detailed Solution|'''<u>Detailed Solution</u>''']]
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!Foundations: &nbsp;
 
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|-
 
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| '''1.''' If &nbsp;<math style="vertical-align: -12px">\lim_{x\rightarrow a} g(x)\neq 0,</math>&nbsp; we have
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{x\rightarrow a} \frac{f(x)}{g(x)}=\frac{\displaystyle{\lim_{x\rightarrow a} f(x)}}{\displaystyle{\lim_{x\rightarrow a} g(x)}}.</math>
 
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| '''2.''' Recall
 
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|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -14px">\lim_{x\rightarrow 0} \frac{\sin x}{x}=1</math>
 
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'''Solution:'''
 
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'''(a)'''
 
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!Step 1: &nbsp;
 
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|-
 
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|Since &nbsp;<math style="vertical-align: -12px">\lim_{x\rightarrow 2} x =2\ne 0,</math>
 
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|we have
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 
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\displaystyle{5} & = & \displaystyle{\lim _{x\rightarrow 2} \bigg[\frac{4-g(x)}{x}\bigg]}\\
 
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&&\\
 
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& = & \displaystyle{\frac{\lim_{x\rightarrow 2} (4-g(x))}{\lim_{x\rightarrow 2} x}}\\
 
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&&\\
 
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& = & \displaystyle{\frac{\lim_{x\rightarrow 2} (4-g(x))}{2}.}
 
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\end{array}</math>
 
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!Step 2: &nbsp;
 
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|-
 
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|If we multiply both sides of the last equation by &nbsp;<math>2,</math>&nbsp; we get
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>10=\lim_{x\rightarrow 2} (4-g(x)).</math>
 
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|Now, using linearity properties of limits, we have
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 
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\displaystyle{10} & = & \displaystyle{\lim_{x\rightarrow 2} 4 -\lim_{x\rightarrow 2}g(x)}\\
 
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&&\\
 
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& = & \displaystyle{4-\lim_{x\rightarrow 2} g(x).}\\
 
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\end{array}</math>
 
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!Step 3: &nbsp;
 
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|-
 
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|Solving for &nbsp;<math style="vertical-align: -12px">\lim_{x\rightarrow 2} g(x)</math>&nbsp; in the last equation,
 
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|we get
 
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&nbsp; &nbsp; &nbsp; &nbsp; <math> \lim_{x\rightarrow 2} g(x)=-6.</math>
 
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'''(b)'''
 
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!Step 1: &nbsp;
 
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|-
 
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|First, we write
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{x\rightarrow 0} \frac{\sin(4x)}{5x}=\lim_{x\rightarrow 0} \frac{4}{5} \bigg(\frac{\sin(4x)}{4x}\bigg).</math>
 
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!Step 2: &nbsp;
 
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|Now, we have
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 
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\displaystyle{\lim_{x\rightarrow 0} \frac{\sin(4x)}{5x}} & = & \displaystyle{\frac{4}{5}\lim_{x\rightarrow 0} \frac{\sin(4x)}{4x}}\\
 
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&&\\
 
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& = & \displaystyle{\frac{4}{5}(1)}\\
 
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&&\\
 
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& = & \displaystyle{\frac{4}{5}.}
 
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\end{array}</math>
 
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|}
 
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'''(c)'''
 
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!Step 1: &nbsp;
 
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|-
 
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|When we plug in &nbsp;<math style="vertical-align: 0px">-3</math>&nbsp; into &nbsp; <math style="vertical-align: -12px">\frac{x}{x^2-9},</math>
 
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|we get &nbsp; <math style="vertical-align: -12px">\frac{-3}{0}.</math>
 
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|Thus,
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{x\rightarrow -3^+} \frac{x}{x^2-9}</math>
 
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|is either equal to &nbsp;<math style="vertical-align: -1px">\infty</math>&nbsp; or &nbsp;<math style="vertical-align: -1px">-\infty.</math>
 
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!Step 2: &nbsp;
 
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|To figure out which one, we factor the denominator to get
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{x\rightarrow -3^+} \frac{x}{x^2-9}=\lim_{x\rightarrow -3^+} \frac{x}{(x-3)(x+3)}.</math>
 
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|We are taking a right hand limit. So, we are looking at values of &nbsp;<math style="vertical-align: 0px">x</math>
 
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|a little bigger than &nbsp;<math style="vertical-align: 0px">-3.</math>&nbsp; (You can imagine values like &nbsp;<math style="vertical-align: 0px">x=-2.9.</math>&nbsp;)
 
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|For these values, the numerator will be negative. 
 
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|Also, for these values, &nbsp;<math style="vertical-align: 0px">x-3</math>&nbsp; will be negative and &nbsp;<math style="vertical-align: -1px">x+3</math>&nbsp; will be positive.
 
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|Therefore, the denominator will be negative.
 
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|Since both the numerator and denominator will be negative (have the same sign),
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{x\rightarrow -3^+} \frac{x}{x^2-9}=\infty.</math>
 
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!Final Answer: &nbsp;
 
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|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; <math> -6</math>
 
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|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; <math>\frac{4}{5}</math>
 
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|&nbsp; &nbsp; '''(c)''' &nbsp; &nbsp; <math>\infty</math>
 
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[[009A_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]
 
[[009A_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

Latest revision as of 13:38, 8 November 2017

Find the following limits:

(a) Find  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 2} g(x),}   provided 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 2} \bigg[\frac{4-g(x)}{x}\bigg]=5.}

(b) Find  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(4x)}{5x} }

(c) Evaluate  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}{x^2-9} }

Solution


Detailed Solution


Return to Sample Exam

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