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Math Help - Proof of Limit

  1. #1
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    Proof of Limit

    I have no experience in proofs whatsoever. Are there any general steps to go about proving math problems? I really don't know how to start these things or what the next steps are after the first step. Here is one from my calculus class I don't know how to do.

    Prove
    \lim_{n\rightarrow 0^+}\sqrt{x} [1+sin^2(2\pi/x)]=0
    Last edited by c_323_h; June 9th 2006 at 12:50 PM.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by c_323_h
    I have no experience in proofs whatsoever. Are there any general steps to go about proving math problems? I really don't know how to start these things or what the next steps are after the first step. Here is one from my calculus class I don't know how to do.

    Prove
    \lim_{n\rightarrow 0^+}\sqrt{x} [1+sin^2(2\pi/x)]=0
    I don't know how formal a proof you want, here is a relativle informal
    proof:

    As for all t \in \mathbb{R} 0 \le |\sin(t)| \le 1 we have that:

    <br />
1 \le 1+\sin^2(2\pi/x) \le 2<br />

    So for all x \ge 0:

    \sqrt{x} \le \sqrt{x} [1+sin^2(2\pi/x)] \le 2\sqrt{x}.

    Now since \sqrt{x} \to 0 as x \to 0 from above this tell us that \sqrt{x} [1+sin^2(2\pi/x)]
    is trapped between two function of x both of which go to 0 as x goes to
    0 from above and hence its limit is also 0.

    RonL
    Last edited by CaptainBlack; June 9th 2006 at 11:41 PM.
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