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

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!Foundations:    
 
!Foundations:    
 
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|Review L'Hopital's Rule
 
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!Step 1:    
 
!Step 1:    
 
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|We begin by factoring the numerator. We have
 
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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>.
 
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|So, we can cancel <math>x+3</math> in the numerator and denominator. Thus, we have
 
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|<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;
 
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|Now, we can just plug in <math>x=-3</math> to get
 
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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>.
 
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!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
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|'''(a)'''
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|'''(a)''' <math>9</math>.
 
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|'''(b)'''  
 
|'''(b)'''  

Revision as of 17:16, 14 February 2016

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:  
Review L'Hopital's Rule

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:  
Step 2:  
Step 3:  

(c)

Step 1:  
Step 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)
(c)

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