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lucy4104

  • 2 years ago

is lim as x approaches c the same as f(c)?

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  1. ghazi
    • 2 years ago
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    where is the function f(c)

  2. lucy4104
    • 2 years ago
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    c=3, f(x)= (x^2+5)/(x-6) continuous or no?... is the question.

  3. ByteMe
    • 2 years ago
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    at c=3, yes... the function is defined at x=3 so the limit of f is f(3)

  4. lucy4104
    • 2 years ago
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    what if x=c is not defined?

  5. ByteMe
    • 2 years ago
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    if f(x) is not defined at x=c, then the limit can still exist but not necessarily at f(c). also, if there is a vertical asymptote at x=c, then the limit does not exist.

  6. lucy4104
    • 2 years ago
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    the limit can still exist but not necessarily at f(c)?

  7. ByteMe
    • 2 years ago
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    yes... for example... |dw:1361168628388:dw| here, the limit of f(x) as x approaches c is a.... NOT f(c)

  8. lucy4104
    • 2 years ago
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    OHHH so then in that case it would be dicontinuous? When can you tell, or what can you do, to know that something is continuous, but algebraically? step by step and explain?

  9. ByteMe
    • 2 years ago
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    yes... but in your original function, it is continuous at all x values except at x=6. so in your problem, the limit of f as x approaches c for any value OTHER THAN 6, will be f(c)

  10. lucy4104
    • 2 years ago
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    oh! so then if x=6 was not an asymptote, but a hole, the limit would be equivalent to f(c)??

  11. ByteMe
    • 2 years ago
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    ***IF*** the function was CONTINUOUS at x=6, then the limit would be f(6)

  12. ByteMe
    • 2 years ago
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    seems like we're going in circles here but we're not... Continuity of a funtion is defined in terms of limits....

  13. lucy4104
    • 2 years ago
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    so...no?

  14. ByteMe
    • 2 years ago
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    if the "hole" you're referring to is a REMOVABLE discontinuity, then yes, the limit would be f(6) or as you said, f(c)...

  15. lucy4104
    • 2 years ago
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    yes!!!! ok, thank you!

  16. ByteMe
    • 2 years ago
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    yw...:)

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