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Math Help - find the limit of the given sequence.

  1. #1
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    find the limit of the given sequence.

    If the limit exists, find the limit of the sequence:

    heres the equation: http://img402.imageshack.us/img402/5907/untitledav2.jpg

    any help with this one please.
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  2. #2
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    Observe the inequalities,
    \ln (1+e^n)\geq \ln (e^n)=n
    \ln (1+e^n)\leq \ln (e^n+e^n)=\ln (2e^n)=\ln 2+\ln e^n=n+\ln 2

    Thus,
    \frac{1}{\ln (1+e^n)} \leq \frac{1}{n}
    \frac{1}{\ln(1+e^n)} \geq \frac{1}{n+\ln 2}
    Thus, multiply by n>0,
    \frac{n}{n+\ln 2}\leq \frac{n}{\ln (1+e^n)}\leq \frac{n}{n}=1

    Now,
    \lim_{n\to \infty} \frac{n}{n+\ln 2}=1
    \lim_{n\to \infty} 1=1
    Thus, by the squeeze theorem,
    \lim_{n\to \infty} \frac{n}{\ln (1+e^n)}=1
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  3. #3
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    okay, i see that we used the squeeze theorem, however,

    in the second part of 'Observe the inequalities' where it = n + ln2

    why did we raise 1 to e^n?

    also, we used 1/n for a bound in the squeeze theorem, is this because you recognized it as a series? if so, which one.

    thanks for all the help!
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  4. #4
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    Quote Originally Posted by rcmango View Post
    okay, i see that we used the squeeze theorem, however,

    in the second part of 'Observe the inequalities' where it = n + ln2

    why did we raise 1 to e^n?

    also, we used 1/n for a bound in the squeeze theorem, is this because you recognized it as a series? if so, which one.

    thanks for all the help!
    Because,
    1\leq e^n
    Thus,
    1+e^n\leq e^n+e^n=2e^n
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  5. #5
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    okay, i can see as both sides of the squeeze theorem converge to 1.
    then, also the middle equation converges to 1.

    thanks for your help.
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