x approaches 0 through positive values, so what's the point of the absolute value of x ?
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:
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
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 but I am not sure where to go from there!
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.
To show without using its power series expansion, we can use the Mean Value Theorem several times.
Let . Then , , and . Using a little trigonometry, we find for all .
Choose any . By the Mean Value Theorem, there exists such that . Again, by the Mean Value Theorem, there exists such that . For the third time, by the Mean Value Theorem, there exists such that . Since , we have . Since , we have the product of four positive numbers is positive, so for any .
This tells us for all .