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owlet

  • one year ago

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  1. owlet
    • one year ago
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  2. owlet
    • one year ago
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    @SolomonZelman @Loser66 @triciaal

  3. freckles
    • one year ago
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    Have you looked at a graph of f(x)? Have you plugged in x=0? or x=-4? Have you looked at limits as we approached these numbers?

  4. owlet
    • one year ago
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    we're not allowed to use graphing devices. the answer is, 0 is the V.A. but -4 is not a V.A. but one of the zeros in the denominator is -4, so it should be a V.A. this is the part i'm not sure

  5. freckles
    • one year ago
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    at x=-4 you should get 0/0 when you plug in the -4 which means it could be a hole at x=-4 at x=0 you should get -2/0 which tells us we have a va at x=0 if you ever get something/0 where the top something isn't 0 then yep it is a va but if you get 0/0 you have more work

  6. owlet
    • one year ago
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    how to prove it using limits?

  7. freckles
    • one year ago
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    cot(-4) clearly exists but the problem is the other function evaluated at x=-4 \[\lim_{x \rightarrow -4} \frac{\sqrt{x^2+9}-5}{x^2+5x+4} \cdot \frac{\sqrt{x^2+9}+5}{\sqrt{x^2+9}+5}\] try evaluating this limit notice I multiplied that one function by a fancy one

  8. owlet
    • one year ago
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    okay. I get that part now. thanks.. how about the 0?

  9. freckles
    • one year ago
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    you should see this limit exists which tells us we don't have a va

  10. freckles
    • one year ago
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    at x=-4

  11. freckles
    • one year ago
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    anyways if you plug in x=0 you get something/0 where the top something isn't 0

  12. freckles
    • one year ago
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    this means you have a va at x=0

  13. freckles
    • one year ago
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    consider evaluating this limit if you are not sure \[\lim_{x \rightarrow 0}\cot(x)\]

  14. Loser66
    • one year ago
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    I think if you just multiple the term as what freckles did, and then simplify, x-4 no more in denominator. Hence, no need to take the lim

  15. Loser66
    • one year ago
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    And that is enough to conclude that x =4 is not a V.A

  16. freckles
    • one year ago
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    (x+4) ? and x=-4? @Loser66 :p

  17. Loser66
    • one year ago
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    Yes, hehehe.. sorry.

  18. owlet
    • one year ago
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    ok thank you both :)

  19. Loser66
    • one year ago
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    |dw:1443637036239:dw|

  20. Loser66
    • one year ago
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    |dw:1443637115972:dw|

  21. freckles
    • one year ago
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    by the way if a limit does not exist the answer doesn't always say the discontinuity is a vertical asymptote we just knew our function didn't have any jump discontinuities

  22. freckles
    • one year ago
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    more details: \[\lim_{x \rightarrow a^{\pm}} f(x)=\pm \infty \text{ then } va \text{ at } x=a \\\]

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