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Math Help - Given certain conditions, find limit?

  1. #1
    Super Member fardeen_gen's Avatar
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    Given certain conditions, find limit?

    If f : [-1, 1] \rightarrow \mathbb{R} and f'(0) = \lim_{n\rightarrow \infty} nf\left(\frac{1}{n}\right) and f(0) = 0, find the value of \lim_{n\rightarrow \infty} \left\{\frac{2}{\pi}(n + 1)\arccos \left(\frac{1}{n}\right) - n\right\}.
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  2. #2
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    Poorly worded question

    Define f(x) such that nf(\frac1n)=\frac2\pi(n+1)\arccos(\frac1n)-n, so f(x)=\frac2\pi(x+1)\arccos x -1 - checking, f(0)=0, as required. So f'(x)=\frac2\pi\left[\arccos x-(x+1)\frac1{\sqrt{1-x^2}}\right], making f'(0)=1-\frac2\pi
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  3. #3
    Lord of certain Rings
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    Quote Originally Posted by fardeen_gen View Post
    If f : [-1, 1] \rightarrow \mathbb{R} and f'(0) = \lim_{n\rightarrow \infty} nf\left(\frac{1}{n}\right) and f(0) = 0, find the value of \lim_{n\rightarrow \infty} \left\{\frac{2}{\pi}(n + 1)\arccos \left(\frac{1}{n}\right) - n\right\}.
    What has the "if" conditions got to do with the limit?

    There is no f(x) in the limit!
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  4. #4
    Senior Member
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    Theorem?

    I think the problem should read: If f : [-1, 1] \rightarrow \mathbb{R} and f(0) = 0, then f'(0) = \lim_{n\rightarrow \infty} nf\left(\frac{1}{n}\right) . Use this theorem to evaluate the following limit...

    This "theorem" makes sense because f'(0)=\lim_{h\to0}\frac{f(0+h)-f(0)}{h-0}=\lim_{h\to0}\frac1h f(h)=\lim_{n\to\infty}nf(\frac1n)

    It was just a poorly worded question, but an interesting technique I had not seen before.
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