1. ## 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:
$\displaystyle \lim_{x \to 0^+} \frac{x - sin(|x|)}{x^3} = \frac{1}{6}$

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

2. ## Re: Challenging limit

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

3. ## 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 $\displaystyle \frac{x- sin(x)}{x^3}= \frac{1}{x^2}\left(1- \frac{sin(x)}{x}\right)$ but I am not sure where to go from there!

4. ## Re: Challenging limit

starting with the Cusa-Huygens inequality

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

we can show that

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

then divide by $\displaystyle x^2$ and take the limit as x -> 0

5. ## Re: Challenging limit

Originally Posted by Idea
starting with the Cusa-Huygens inequality

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

we can show that

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

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

-Dan

6. ## 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.

7. ## Re: Challenging limit

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

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

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

This tells us $\displaystyle x^3-6x+6\sin x > 0 \Rightarrow x^3 > 6x-6\sin x \Rightarrow \dfrac{x^2}{6} > 1-\dfrac{\sin x}{x}$ for all $\displaystyle 0 < x < 2\pi$.

8. ## Re: Challenging limit

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

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

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

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

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