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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|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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|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  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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