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Math Help - sequence, find the limit.

  1. #1
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    sequence, find the limit.

    heres the equation for this sequence: http://img401.imageshack.us/img401/6712/untitledfe2.jpg

    i'm asked to find the limit. I thought i could use the dqueeze theorem, i may have been on track using 1/sqrt(n) ?

    help need please.
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  2. #2
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    Note that,
    -1\leq \sin x \leq 1
    Adding 1,
    0\leq 1+\sin x\leq 2
    Divide by 1+\sqrt{n}>0
    Thus,
    0=\frac{0}{1+\sqrt{n}}\leq \frac{1+\sin x}{1+\sqrt{n}}\leq \frac{2}{1+\sqrt{n}}

    The limits,
    \lim_{n \to \infty} 0 =0

    \lim_{n\to \infty} \frac{2}{1+\sqrt{n}}=0

    Thus,
    \lim_{n\to \infty} \frac{1+\sin x}{1+\sqrt{n}}=0
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  3. #3
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    i understand. thanks.

    also, where you added one. you state +sin because u also added +1 to the middle term also, correct?

    thanks alot, excellent.
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