Difference between revisions of "Math 22 Asymptotes"
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|Therefore, <math>\lim_{x\to 3^-}\frac{-2}{x-3}=\frac{\text{negative}}{\text{negative}}=\infty</math> | |Therefore, <math>\lim_{x\to 3^-}\frac{-2}{x-3}=\frac{\text{negative}}{\text{negative}}=\infty</math> | ||
+ | |} | ||
+ | |||
+ | '''2)''' <math>\lim_{x\to 2^+}\frac{-5}{x-2}</math> | ||
+ | {| class = "mw-collapsible mw-collapsed" style = "text-align:left;" | ||
+ | !Solution: | ||
+ | |- | ||
+ | |Notice <math>x\to 2^+</math>, so <math> x>2 </math>, then <math> x-2>0</math>, hence the denominator will be "positive". | ||
+ | |- | ||
+ | |Therefore, <math>\lim_{x\to 2^+}\frac{-5}{x-2}=\frac{\text{negative}}{\text{positive}}=-\infty</math> | ||
|} | |} | ||
This page is under construction | This page is under construction |
Revision as of 06:56, 4 August 2020
Vertical Asymptotes and Infinite Limits
If approaches infinity (or negative infinity) as approaches from the right or from the left, then the line
Example: Find the limit below:
1)
Solution: |
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Important: is either or |
Notice , so , then , hence the denominator will be "negative". |
Therefore, |
2)
Solution: |
---|
Notice , so , then , hence the denominator will be "positive". |
Therefore, |
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This page were made by Tri Phan