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lornbeach Group Title

Find the limit of the function algebraically. lim(x -> 5) x^2-25/x-5

  • 4 months ago
  • 4 months ago

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  1. myininaya Group Title
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    So I think you mean (x^2-25)/(x-5) right? This expression does not exist at x=5 because when you plug in 5 you get something/0 Now when you have 0/0 that means the limit can exist Is there any algebra you can perform to force a function that is continuous at 5?

    • 4 months ago
  2. myininaya Group Title
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    Basically is there any factoring you can do?

    • 4 months ago
  3. lornbeach Group Title
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    Yes that's it and simplify iit

    • 4 months ago
  4. myininaya Group Title
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    yep then plug in 5 to resulting function that is continuous at 5

    • 4 months ago
  5. myininaya Group Title
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    you can share your steps with me if you want

    • 4 months ago
  6. lornbeach Group Title
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    would it be 10/0?

    • 4 months ago
  7. myininaya Group Title
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    not 10/0 nope show me what you did to get that

    • 4 months ago
  8. lornbeach Group Title
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    (x+25)(x-25)/x-5

    • 4 months ago
  9. myininaya Group Title
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    that isn't how you factor x^2-25

    • 4 months ago
  10. myininaya Group Title
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    x^2-a^2=(x-a)(x+a) you have x^2-5^2=(x-5)(x+5)

    • 4 months ago
  11. lornbeach Group Title
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    OH (x+5)(x-5)

    • 4 months ago
  12. myininaya Group Title
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    right but what do yo notice about the top and bottom (As far as factors go)

    • 4 months ago
  13. myininaya Group Title
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    and you have (x+5)(x-5)/(x-5)

    • 4 months ago
  14. lornbeach Group Title
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    You can cancel

    • 4 months ago
  15. lornbeach Group Title
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    (x+5)

    • 4 months ago
  16. myininaya Group Title
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    right you basically get rid of the discontinuity so you can plug in x=5

    • 4 months ago
  17. lornbeach Group Title
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    so 10

    • 4 months ago
  18. myininaya Group Title
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    right

    • 4 months ago
  19. lornbeach Group Title
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    Thank you!!

    • 4 months ago
  20. myininaya Group Title
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    np

    • 4 months ago
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