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Math Help - L'Hospital's Rule

  1. #1
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    L'Hospital's Rule

    Find the limit of x(tan(8/x)) as it approaches infinity.

    If I plug in infinity, I get (infinity * 0) so the problem is an indeterminate form.

    So I find the limit of tan(8/x) / (1/x). This is (0 * 0) so I can use l'Hospital's rule:

    sec^2(8/x)(-8x^-2) / 1

    That still is (0 * 0), so I can use l'Hospital's again. I get:

    (16x^-3) (2sec(8/x)(-8x^-2)

    At this time I realize that this doesn't look like it's helping me, so I probably did something wrong. Can someone look over my work, tell me if I'm on the right track, and if not, tell me what I did wrong? Thanks.
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  2. #2
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    Quote Originally Posted by Maziana View Post
    Find the limit of x(tan(8/x)) as it approaches infinity.

    If I plug in infinity, I get (infinity * 0) so the problem is an indeterminate form.

    So I find the limit of tan(8/x) / (1/x). This is (0 * 0) so I can use l'Hospital's rule:

    sec^2(8/x)(-8x^-2) / 1

    That still is (0 * 0), so I can use l'Hospital's again. I get:

    (16x^-3) (2sec(8/x)(-8x^-2)

    At this time I realize that this doesn't look like it's helping me, so I probably did something wrong. Can someone look over my work, tell me if I'm on the right track, and if not, tell me what I did wrong? Thanks.
    Your derivative of \tan \left( \frac{8}{x}\right) is wrong. You need to use the chain rule.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    Your derivative of \tan \left( \frac{8}{x}\right) is wrong. You need to use the chain rule.
    Actually the derivative of \tan{\left(\frac{8}{x}\right)}, which is \sec^2{\left(\frac{8}{x}\right)}[-8x^{-2}] is correct.

    It is the derivative of \frac{1}{x} in the denominator that is incorrect.
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  4. #4
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    That's right; thanks. 1/x should go to -1/x^2. So it should be:

    sec^2(8/x)(-8x^-2) / -1/x^2

    That's still (0 * 0), so I can use l'Hospital's again. But I still don't see how that helps; it just makes the problem even more complicated.
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  5. #5
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    Try setting t=1/x now when x\to\infty, t\to0

    Then you can calculate the limit:

    \lim\limits_{t\to0}\frac{\tan(8t)}{t}.
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