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anonymous

  • one year ago

GIVING OUT MEDALS N FAN 3.)Find the limit of the function by using direct substitution. lim x-->1 (x^2+3x-1) a) 0 b) does not exist c) 3 d) -3 4.) Find the limit of the function algebraically lim x--> (x^2-100)/(x+10) a)-10 b)-20 c) does not exist d) 1

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  1. anonymous
    • one year ago
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    @welshfella

  2. Astrophysics
    • one year ago
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    @em2000 what did you get?

  3. anonymous
    • one year ago
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    3) b 4) c

  4. Astrophysics
    • one year ago
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    \[\lim_{x \rightarrow 1} (x^2+3x-1) \implies ((1)^2+3(1)-1)\]

  5. Astrophysics
    • one year ago
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    Why not direct sub as it tells you to?

  6. Astrophysics
    • one year ago
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    \[(x^2-100) \implies (x-10)(x+10)\]

  7. Astrophysics
    • one year ago
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    You should be able to do the rest

  8. Australopithecus
    • one year ago
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    Yeah as he said \[\lim_{x \rightarrow 1} x^2 + 3x - 1 = 1^2 + 3*1 - 1\]

  9. anonymous
    • one year ago
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    oh ok so 3) is c

  10. Astrophysics
    • one year ago
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    Yeah, and what is the limit for 4?

  11. anonymous
    • one year ago
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    -10

  12. Astrophysics
    • one year ago
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    As in what is x approaching you have not put that

  13. anonymous
    • one year ago
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    ik it is lim x--> -10 ( i forgot to put the -10)

  14. Astrophysics
    • one year ago
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    Oh ok, so you should be able to do an easy cancellation and your answer will be?

  15. Australopithecus
    • one year ago
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    You need to use algebra to solve this because if you directly sub in 10 you will get a divide by zero

  16. Australopithecus
    • one year ago
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    you cannot divide by zero in mathematics

  17. Astrophysics
    • one year ago
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    \[\lim_{x \rightarrow -10} \frac{ (x-10)(x+10) }{ (x+10) }\]

  18. anonymous
    • one year ago
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    so it does not exist

  19. Australopithecus
    • one year ago
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    It exists, you just need to use algebra to solve it

  20. Astrophysics
    • one year ago
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    Why do you say that? Do you see something that's getting cancelled

  21. Australopithecus
    • one year ago
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    Astrophysics showed you the factored form of the top

  22. anonymous
    • one year ago
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    what did u get>

  23. anonymous
    • one year ago
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    -20

  24. Astrophysics
    • one year ago
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    Yes.

  25. anonymous
    • one year ago
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    sry bout that

  26. Australopithecus
    • one year ago
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    Example: \[\frac{(x+1)(x+2)}{(x+1)} = (x+2)*1\]

  27. anonymous
    • one year ago
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    quick one, for this i got does not exist is that right? Find the limit of the function algebraically. lim x-->0 (x^2 - 2x)/(x^4) a) does not exist b) 8 c) 0 d) -8

  28. Astrophysics
    • one year ago
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    Show your work and explain why

  29. anonymous
    • one year ago
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    wait its 0

  30. Astrophysics
    • one year ago
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    Well that's not very useful, you're just guessing it seems, once you figure out the reason please do share!

  31. anonymous
    • one year ago
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    (0^2 - 0x)/(0^4)= 0/0=0

  32. Astrophysics
    • one year ago
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    No that's not how it works, try seeing if the limit exists by taking the right and left hand side limits.

  33. anonymous
    • one year ago
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    what

  34. Astrophysics
    • one year ago
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    Well I don't have the time to teach the concepts of limits but I suggest you watch this video and others related https://www.youtube.com/watch?v=riXcZT2ICjA

  35. Australopithecus
    • one year ago
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    You cannot divide by zero, so if you see zero as the denominator after doing a substitution you need to either solve algebraically or test both sides of the limit

  36. Australopithecus
    • one year ago
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    What I mean by both sides is that you take a limit using an extremely small number from the left we will denote it: \[0^-\] This would be like -0.00000000000000000000006 or something so small its almost zero but not quite you do the same for the other side (right side) \[0^+\] This will be 0.0000000000000000000000003 or something so small its almost zero but not quite If \[\lim_{x \rightarrow 0^-} f(x) = \lim_{x \rightarrow 0^+} f(x)\] The limit is said to exist. If they do not equal the limit does not exist Note that being said, when a number is extremely tiny and in the denominator your function is going to equal infinity

  37. Australopithecus
    • one year ago
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    See example 1 and maybe read through his notes http://tutorial.math.lamar.edu/Classes/CalcI/OneSidedLimits.aspx

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