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Math Help - Challenging limit

  1. #1
    Forum Admin topsquark's Avatar
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    Challenging limit

    I came up with a nice little limit for the Challenge Problems section. Unfortunately I can't actually solve the thing, so it's not a candidate. But it's been driving me crazy. Here it is:
    \lim_{x \to 0^+} \frac{x - sin(|x|)}{x^3} = \frac{1}{6}

    and, of course, no l'Hopital's or series expansions. I even tried Wolfram|Alpha and it did it by l'Hopital's.

    I've got two approaches which might bear fruit but I don't quite know how to apply them. The first is using the squeeze theorem, and the second is a good old fashioned epsilon-delta method. The problem here is that the function is discontinuous at the origin and I can't see how to apply either method.

    Thanks!

    -Dan
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    Re: Challenging limit

    x approaches 0 through positive values, so what's the point of the absolute value of x ?
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    Re: Challenging limit

    As "Idea" (nice name!) said, since x is always positive, sin(|x|)= sin(x). I would then replace sin(x) by its expression as a MacLaurin series: x- x^3/6+ x^5/5!- ... so that x- sin(|x|)= x- sin(x)= x^3/6- x^5/5!+ ...

    Okay, I now see that you say "No L'Hopital or series expansion"! So instead, write \frac{x- sin(x)}{x^3}= \frac{1}{x^2}\left(1- \frac{sin(x)}{x}\right) but I am not sure where to go from there!
    Last edited by HallsofIvy; June 10th 2014 at 06:59 AM.
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    Re: Challenging limit

    starting with the Cusa-Huygens inequality

    \frac{\sin  x}{x}<\frac{2+\cos  x}{3}\text{   }  ( 0 < x < \pi /2 )

    we can show that

    \frac{\sin ^2x}{3(1+\cos  x)}\leq  1-\frac{\sin  x}{x}\leq  \frac{x^2}{6}

    then divide by x^2 and take the limit as x -> 0
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  5. #5
    Forum Admin topsquark's Avatar
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    Re: Challenging limit

    Quote Originally Posted by Idea View Post
    starting with the Cusa-Huygens inequality

    \frac{\sin  x}{x}<\frac{2+\cos  x}{3}\text{   }  ( 0 < x < \pi /2 )

    we can show that

    \frac{\sin ^2x}{3(1+\cos  x)}\leq  1-\frac{\sin  x}{x}\leq  \frac{x^2}{6}

    then divide by x^2 and take the limit as x -> 0
    Ah, that's better. Thanks to you both.

    -Dan
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    Re: Challenging limit

    I am taking nothing away from idea, whose proof seems quite elegant. But there are a few places where I cannot follow it.

    Problem 1. These may be typos. I can get $\dfrac{sin^2(x)}{3(1 + cos(x))} < 1 - \dfrac{sin(x)} < \dfrac{x^2}{6} (0 < x < 1),$

    but not $\dfrac{sin^2(x)}{3(1 + cos(x))} \le 1 - \dfrac{sin(x)}{x} \le \dfrac{x^2}{6} (0 < x < \pi / 2).$

    Of course less than implies less than or equal, but the $\dfrac{\pi }{2}$ has me completely stumped on $1 - \dfrac{sin(x)}{x} \le \dfrac{x^2}{6}.$

    Problem 2. As for the $1 - \dfrac{sin(x)}{x} < \dfrac{x^2}{6},$ I can only get that from a series expansion, not from $\dfrac{sin(x)}{x} < \dfrac{2 + cos(x)}{3}.$

    Excuse me if my questions have obvious answers.
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    Re: Challenging limit

    To show 1-\dfrac{\sin x}{x} \le \dfrac{x^2}{6} without using its power series expansion, we can use the Mean Value Theorem several times.

    Let f(x) = x^3-6x+6\sin x. Then f'(x) = 3x^2-6+6\cos x, f''(x) = 6x-6\sin x, and f'''(x) = 6-6\cos x. Using a little trigonometry, we find f'''(x) > 0 for all 0 < x < 2\pi.

    Choose any 0 < x < 2\pi. By the Mean Value Theorem, there exists c_1\in (0,x) such that f(x) - f(0) = (x-0)f'(c_1). Again, by the Mean Value Theorem, there exists c_2 \in (0,c_1) such that f'(c_1)-f'(0) = (c_1-0)f''(c_2). For the third time, by the Mean Value Theorem, there exists c_3\in (0,c_2) such that f''(c_2)-f''(0) = (c_2-0)f'''(c_3). Since f(0)=f'(0)=f''(0)=0, we have f(x) = xf'(c_1) = xc_1f''(c_2) = xc_1c_2f'''(c_3). Since 0<c_3<c_2<c_1<x, we have the product of four positive numbers is positive, so f(x)>0 for any 0<x<2\pi.

    This tells us x^3-6x+6\sin x > 0 \Rightarrow x^3 > 6x-6\sin x \Rightarrow \dfrac{x^2}{6} > 1-\dfrac{\sin x}{x} for all 0 < x < 2\pi.
    Last edited by SlipEternal; June 11th 2014 at 11:08 AM.
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  8. #8
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    Re: Challenging limit

    Quote Originally Posted by SlipEternal View Post
    To show 1-\dfrac{\sin x}{x} \le \dfrac{x^2}{6} without using its power series expansion, we can use the Mean Value Theorem several times.

    Let f(x) = x^3-6x+6\sin x. Then f'(x) = 3x^2-6+6\cos x, f''(x) = 6x-6\sin x, and f'''(x) = 6-6\cos x. Using a little trigonometry, we find f'''(x) > 0 for all 0 < x < 2\pi.

    Choose any 0 < x < 2\pi. By the Mean Value Theorem, there exists c_1\in (0,x) such that f(x) - f(0) = (x-0)f'(c_1). Again, by the Mean Value Theorem, there exists c_2 \in (0,c_1) such that f'(c_1)-f'(0) = (c_1-0)f''(c_2). For the third time, by the Mean Value Theorem, there exists c_3\in (0,c_2) such that f''(c_2)-f''(0) = (c_2-0)f'''(c_3). Since f(0)=f'(0)=f''(0)=0, we have f(x) = xf'(c_1) = xc_1f''(c_2) = xc_1c_2f'''(c_3). Since 0<c_3<c_2<c_1<x, we have the product of four positive numbers is positive, so f(x)>0 for any 0<x<2\pi.

    This tells us x^3-6x+6\sin x > 0 \Rightarrow x^3 > 6x-6\sin x \Rightarrow \dfrac{x^2}{6} > 1-\dfrac{\sin x}{x} for all 0 < x < 2\pi.
    Wow, very elegant. Quite clear now.

    I can't tell you how long I wrestled with that. Thanks for the help.
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