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Math Help - Sequence Limits

  1. #1
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    Sequence Limits

    This is the same as finding regular limits, but only using integers, right?

    The problem is: a_n=[ln(2+e^n)]/3n

    How do I solve it?
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  2. #2
    Senior Member apcalculus's Avatar
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    Quote Originally Posted by schnek View Post
    This is the same as finding regular limits, but only using integers, right?

    The problem is: a_n=[ln(2+e^n)]/3n

    How do I solve it?

    You could use L'Hopital's Rule here. Observe that both numerator and denominator go to infinity as n goes to infinity.

    Then the ratio will approach the ratio of the derivatives:

    \frac{e^n}{(2+e^n)3} \rightarrow \frac{1}{3}
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by schnek View Post
    This is the same as finding regular limits, but only using integers, right?

    The problem is: a_n=[ln(2+e^n)]/3n

    How do I solve it?
    Or merely note that \frac{\ln\left(2+e^n\right)}{3n}=\frac{\ln\left(e^  n\left(\frac{2}{e^n}+1\right)\right)}{3n}=\frac{\l  n\left(e^n\right)}{3n}+\frac{\ln\left(\frac{2}{e^n  }+1\right)}{n}=\frac{1}{3}+\frac{\ln\left(\frac{2}  {e^n}+1\right)}{n}. The second term in a senes "doubly" tends to zero. The numerator tends towards \ln\left(1\right)=0 and the denominator is getting unboundedly large.
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