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recon14193

  • 9 months ago

limit evaluation

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  1. recon14193
    • 9 months ago
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    \[\lim_{x \rightarrow -4}\frac{ \sqrt{x^9+9}-5 }{ x+4 }\]

  2. myininaya
    • 9 months ago
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    Try rationalizing the top by multiply the numerator and denominator by the top's conjugate

  3. recon14193
    • 9 months ago
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    I worked it out to \[\frac{ x^9-16 }{ x+4(\sqrt{x^9+9}+5) }\]

  4. recon14193
    • 9 months ago
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    I dont know what I would expand the top to

  5. recon14193
    • 9 months ago
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    @myininaya

  6. myininaya
    • 9 months ago
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    and i think you mean to have parenthesis around the (x+4) too on bottom

  7. recon14193
    • 9 months ago
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    yes

  8. myininaya
    • 9 months ago
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    try factoring the top to see if there is a factor of x+4 in it so you can get rid of what would make the bottom 0

  9. recon14193
    • 9 months ago
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    I dont know how I would factor the top though and also with it being ^9 it's still a negative square root

  10. myininaya
    • 9 months ago
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    do you know how do synthetic division or long division?

  11. recon14193
    • 9 months ago
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    I should but I dont remember

  12. myininaya
    • 9 months ago
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    ok well if you did know how and you determine there was no remainder then the limit does exist but if there is a remainder then the limit will not exist but i guess you can also determine this from just pluggin in -4 on both top and bottom if you have 0/0 the limit will exist if you don't then it won't

  13. myininaya
    • 9 months ago
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    if the limit does exist then you of course must find it but if the limit doesn't exist, then you are done and dne is your answer

  14. recon14193
    • 9 months ago
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    the limit does exist I know that, and I went back to my algebra 2 notes and doing synthetic division (x+4) is not a factor

  15. myininaya
    • 9 months ago
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    how do you get the limit exists?

  16. recon14193
    • 9 months ago
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    This is part of a review and when I put DNE, I was told I was incorrect

  17. myininaya
    • 9 months ago
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    This question doesn't have a real limit had the function only exist for values where x^9+9 is greater than or equal to 0 anyhow x^9 has to be greater to 0 which means x cannot be negative because if it makes since to say if x is negative then x^9 is definitely more and most negative

  18. myininaya
    • 9 months ago
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    sense not since*

  19. myininaya
    • 9 months ago
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    you can even see a numerical approach will also show nothing exists to the left and right of -4 or at -4 (even though we aren't really concerned what happens at -4)

  20. myininaya
    • 9 months ago
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    http://www.wolframalpha.com/input/?i=y%3D%28sqrt%28x%5E9%2B9%29-5%29%29%2F%28x%2B4%29+real of course there are some negative values the function does exist for after all x^9+9 has to be greater than or equal to 0

  21. myininaya
    • 9 months ago
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    i misphrased it a little above when i just said x^9 has to be greater than or equal to 0 only small negative values will work anyhow is what i mean

  22. myininaya
    • 9 months ago
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    as you can also tell from the graph nothing exists around x=-4

  23. myininaya
    • 9 months ago
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    the limit does not exist we have determine this algebraically, graphically, and numerically

  24. myininaya
    • 9 months ago
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    \[\lim_{x \rightarrow -4}\frac{\sqrt{x^9+9}-5}{x+4}\] just to be certain this was the problem right?

  25. recon14193
    • 9 months ago
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    yes

  26. myininaya
    • 9 months ago
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    so what is needed to convince you this limit does not exist? like which explanation does not make sense?

  27. recon14193
    • 9 months ago
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    It's not that none of the explanations don't make sense. I came to the same conclusion working it myself. It's just that when I plugged in my answer what the test told me was that, that answer is incorrect

  28. myininaya
    • 9 months ago
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    we can try to determine where the written went wrong in writing the test

  29. myininaya
    • 9 months ago
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    give me a second and let me see if i can find out where they went wrong in writing the problem because i bet you that is what happened

  30. myininaya
    • 9 months ago
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    writer* not written

  31. recon14193
    • 9 months ago
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    could I ask you one other problem real quick first

  32. myininaya
    • 9 months ago
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    so if they are saying the limit exist then that means they want some (x+4)'s to cancel... so one second

  33. myininaya
    • 9 months ago
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    ok ask

  34. recon14193
    • 9 months ago
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    http://www.wolframalpha.com/input/?i=%28x%2B1%29%2F%281%2B%281%2F%28x%2B1%29%29%29 For the domain why is it x=/=2 and x=/=-1

  35. recon14193
    • 9 months ago
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    -2*

  36. recon14193
    • 9 months ago
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    I got the -2 but not why -1 also

  37. myininaya
    • 9 months ago
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    do me a favor while i look at that try to answer that same question as if you were answering \[\lim_{x \rightarrow -4}\frac{\sqrt{x^2+9}-5}{x+4}\]

  38. myininaya
    • 9 months ago
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    i think you might get it right if you pretend that is the question

  39. recon14193
    • 9 months ago
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    alright

  40. myininaya
    • 9 months ago
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    let me know and if you get it wrong i'm out of guesses of what they meant

  41. myininaya
    • 9 months ago
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    ok you have a fraction inside that fraction

  42. myininaya
    • 9 months ago
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    that one fraction does not exist at x=-1

  43. myininaya
    • 9 months ago
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    and then after that of course you already knew to find when 1+1/(x+1) is not 0 which is when x=-2

  44. myininaya
    • 9 months ago
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    for example when pluggin in -1 we get \[\frac{1}{1+\frac{1}{-1+1}}=\frac{1}{1+\frac{1}{0}} \] but 1/0 is not a number

  45. recon14193
    • 9 months ago
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    alright so for \[\frac{ \sqrt{x^2+9}-5 }{ x+4 }\] I get\[\frac{ (x-4) }{ \sqrt{x^2+9}+5 }\] which still gives a negative root when I plug in -4

  46. myininaya
    • 9 months ago
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    so 1/(1+1/0)) is still not a number

  47. myininaya
    • 9 months ago
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    no it doesn't

  48. recon14193
    • 9 months ago
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    sorry forgot ^2

  49. recon14193
    • 9 months ago
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    so yeah never mind it gives 0

  50. myininaya
    • 9 months ago
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    :(

  51. myininaya
    • 9 months ago
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    ok you plugged in -4 right?

  52. myininaya
    • 9 months ago
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    \[\frac{-4-4}{\sqrt{(-4)^2+9}+5}\] I think you are getting tired of this problem and you aren't going through the arithmetic slowly enough

  53. myininaya
    • 9 months ago
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    -4-4 is -8

  54. myininaya
    • 9 months ago
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    and can you do the bottom?

  55. recon14193
    • 9 months ago
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    \[\sqrt{16+9}=\sqrt{25}....... 5+5=10\] so 8/10 or 4/5

  56. myininaya
    • 9 months ago
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    don't forget the negative sign from the top

  57. myininaya
    • 9 months ago
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    so -4/5 tell me if that is what they were looking for or not

  58. recon14193
    • 9 months ago
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    I don't know because it doesn't show me the correct answer, but it is something

  59. myininaya
    • 9 months ago
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    you should let your teacher know of the mistake though

  60. myininaya
    • 9 months ago
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    most teachers (or some teachers) are amazed by students who can find there errors

  61. myininaya
    • 9 months ago
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    their*

  62. recon14193
    • 9 months ago
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    I haven't met them yet, and I don't head to school until another week. What would be the most respectable non-annoying way to present it?

  63. myininaya
    • 9 months ago
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    Definitely don't rub in there face. I would just bring it up once if and when they respond. I think an email would be just fine. You can be like: Hey professor, As I was taking the exam, I came across the problem lim (x->-4) (sqrt(x^9+9)-5)/(x+4) and I said the limit does not exist but it was returned incorrect by the exam. I was thinking it was meant to be lim (x->-4) (sqrt(x^2+9)-5)/(x+4) where the answer would be -4/5. Please let me know if I'm correct or not. Sincerely, (you) But don't say she made a mistake. Just say the exam or whatever. What I'm saying is don't say YOU in the email. Try not to imply it is the teacher's fault even though it is. I think that is the best way.

  64. myininaya
    • 9 months ago
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    Or dear professor if you are feeling more formal.

  65. recon14193
    • 9 months ago
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    alright should I provide an explanation of my work about how I could not find the limit algebraically, graphically, or numerically for the original problem

  66. myininaya
    • 9 months ago
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    I think you could say just nothing exists around -4 since x^9+9 has to be greater than or equal to 0. You know which means x^9>=-9 But for x values less than -4 certainly do not satisfy the inequality x^9>=-9. You could actually solve that inequality if you wanted to. x>=(-9)^(1/9) but values really close to -4 are definitely not in the set [(-9)^(1/9),inf)

  67. recon14193
    • 9 months ago
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    alright thank you for all you help

  68. myininaya
    • 9 months ago
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    Np. I don't have problem helping you with this one on your test mainly because it was written wrong. But I hope you intend to give the rest of the test your own knowledge.

  69. myininaya
    • 9 months ago
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    Good luck @recon14193

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