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anonymous

  • 4 years ago

how do i find the limit ln(x-3) x-->3+

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  1. TuringTest
    • 4 years ago
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    it's approaching from the right, so it's defined because x-3 is positive and what number does lny approach as y approaches zero?

  2. lgbasallote
    • 4 years ago
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    ln 0 = infinity

  3. TuringTest
    • 4 years ago
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    ...because x^a is always positive for x>0 we can only make it small by making 'a' large negative

  4. anonymous
    • 4 years ago
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    oh kinda get it

  5. anonymous
    • 4 years ago
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    you migth think of it like this. the log is the inverse of the exponential. when would \[e^x=0\]? well never, but it would be close to zero if x went to minus infinity

  6. TuringTest
    • 4 years ago
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    yes it should be minus^ Igb is a bit off

  7. anonymous
    • 4 years ago
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    so as x goes to minus infinity, \[e^x\] goes to zero and as x goes to zero, \[\ln(x)\] goes to minus infinity

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