Difference between revisions of "Math 22 Asymptotes"

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   If <math>f(x)</math> approaches infinity (or negative infinity) as <math>x</math> approaches <math>c</math> from the right or from the left, then the line
 
   If <math>f(x)</math> approaches infinity (or negative infinity) as <math>x</math> approaches <math>c</math> from the right or from the left, then the line
  
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'''Example''': Find the limit below:
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'''Example''': Find the a vertical Asymptotes as below:
  
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'''1)''' <math>\lim_{x\to 3^-}\frac{-2}{x-3}</math>  
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'''1)''' <math>f(x)=\frac{x+3}{x^2-4}</math>  
 
{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Solution: &nbsp;
 
!Solution: &nbsp;
 
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|-
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|'''Important''': <math>\lim \frac{\text{constant}}{0}</math> is either <math>\infty</math> or <math>-\infty</math>
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|Notice <math>f(x)\frac{x+3}{x^2-4}=\frac{x+3}{(x-2)(x+2)}</math>
 
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|-
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|Notice <math>x\to 3^-</math>, so <math> x<3 </math>, then <math> x-3<0</math>, hence the denominator will be "negative".
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|Let the denominator equals to zero, ie: <math>(x-2)(x+2)=0</math>, hence <math>x=-2</math> or <math>x=2</math>
 
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|-
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|Therefore, <math>\lim_{x\to 3^-}\frac{-2}{x-3}=\frac{\text{negative}}{\text{negative}}=\infty</math>  
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|Therefore, <math>f(x)</math> has vertical asymptotes at <math>x=2</math> and <math>x=-2</math>
 
|}
 
|}
  
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'''2)''' <math>\lim_{x\to 2^+}\frac{-5}{x-2}</math>  
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'''2)''' <math>f(x)=\frac{x^2-x-6}{x^2-9}</math>  
 
{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Solution: &nbsp;
 
!Solution: &nbsp;
 
|-
 
|-
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|Notice <math>x\to 2^+</math>, so <math> x>2 </math>, then <math> x-2>0</math>, hence the denominator will be "positive".
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|Notice <math>f(x)\frac{x^2-x-6}{x^2-9}=\frac{(x-3)(x+2)}{(x-3)(x+3)}=\frac{x+2}{x+3}</math>
 
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|-
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|Therefore, <math>\lim_{x\to 2^+}\frac{-5}{x-2}=\frac{\text{negative}}{\text{positive}}=-\infty</math>  
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|Let the denominator equals to zero, ie: <math>(x+3)=0</math>, hence <math>x=-3</math>
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|-
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|Therefore, <math>f(x)</math> has vertical asymptote at <math>x=-2</math>
 
|}
 
|}
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This page is under construction
 
This page is under construction
  

Revision as of 07:07, 4 August 2020

Vertical Asymptotes and Infinite Limits

 If  approaches infinity (or negative infinity) 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}
 approaches 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 c}
 from the right or from the left, then the line

Example: Find the a vertical Asymptotes as below:

1) 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 f(x)=\frac{x+3}{x^2-4}}

Solution:  
Notice 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 f(x)\frac{x+3}{x^2-4}=\frac{x+3}{(x-2)(x+2)}}
Let the denominator equals to zero, ie: 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-2)(x+2)=0} , hence 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=-2} 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 x=2}
Therefore, 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 f(x)} has vertical asymptotes at 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=2} 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 x=-2}

2) 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 f(x)=\frac{x^2-x-6}{x^2-9}}

Solution:  
Notice 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 f(x)\frac{x^2-x-6}{x^2-9}=\frac{(x-3)(x+2)}{(x-3)(x+3)}=\frac{x+2}{x+3}}
Let the denominator equals to zero, ie: 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)=0} , hence 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}
Therefore, 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 f(x)} has vertical asymptote at 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=-2}

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