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lornbeach

  • 6 months ago

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

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  1. myininaya
    • 6 months ago
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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?

  2. myininaya
    • 6 months ago
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    Basically is there any factoring you can do?

  3. lornbeach
    • 6 months ago
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    Yes that's it and simplify iit

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

  5. myininaya
    • 6 months ago
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    you can share your steps with me if you want

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

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

  8. lornbeach
    • 6 months ago
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    (x+25)(x-25)/x-5

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

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

  11. lornbeach
    • 6 months ago
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    OH (x+5)(x-5)

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

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

  14. lornbeach
    • 6 months ago
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    You can cancel

  15. lornbeach
    • 6 months ago
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    (x+5)

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

  17. lornbeach
    • 6 months ago
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    so 10

  18. myininaya
    • 6 months ago
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    right

  19. lornbeach
    • 6 months ago
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    Thank you!!

  20. myininaya
    • 6 months ago
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    np

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