Difference between revisions of "Math 22 Asymptotes"

Revision as of 07:07, 4 August 2020

Vertical Asymptotes and Infinite Limits

 If  $f(x)$  approaches infinity (or negative infinity) as  $x$  approaches  $c$  from the right or from the left, then the line

Example: Find the a vertical Asymptotes as below:

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

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

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

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

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

Difference between revisions of "Math 22 Asymptotes"

Revision as of 07:07, 4 August 2020

Vertical Asymptotes and Infinite Limits

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