My Math Forum I can't show why x->0 of sin(x)/x is 1!!!

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July 8th, 2019, 05:31 AM   #11
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Quote:
 Originally Posted by babaliaris Can someone explain why I can't use L'Hospital's rule here? Are you sure about it? From what I remember from high school, if you have a limit of the form $\displaystyle \frac{0}{0}$ use can use L'Hospital's rule.
In principle, you can use the rule. But the rule requires you to know the derivative of sin(x). If you do everything really structured, then you need to avoid circularities. That is, you'd need to derive the derivative of sin(x) independently from the limit. That is often a problem. That is: a lot of books prove sin' = cos USING the limit sin(x)/x --->1.

July 8th, 2019, 05:33 AM   #12
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Quote:
 Originally Posted by Micrm@ss In principle, you can use the rule. But the rule requires you to know the derivative of sin(x). If you do everything really structured, then you need to avoid circularities. That is, you'd need to derive the derivative of sin(x) independently from the limit. That is often a problem. That is: a lot of books prove sin' = cos USING the limit sin(x)/x --->1.
Oh I understand now.

 July 8th, 2019, 05:36 AM #13 Senior Member   Joined: Oct 2009 Posts: 841 Thanks: 323 Do note that not all derivations of sin' = cos use the limit sin(x)/x --->1 in a fundamental way. For example, if you define sin using power series, or even as a solution of the differential equation y'' = -y, then you won't really need the limit sin(x)/x--->1. In that case, l'Hospital's rule is perfectly fine. Thanks from babaliaris
July 8th, 2019, 12:50 PM   #14
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 Originally Posted by romsek thank you so much for stopping by to criticize what's been posted. I don't suppose it would be too much to ask that you post what you consider an adequate solution to the problem being asked would it?
Nope.

I thought "This is typically done by a geometric argument which is difficult to carry out without drawing the right picture" was fairly clear. This is a pretty standard geometric argument and I'm not going to spend time recreating this picture to save someone 5 seconds of googling.

 July 8th, 2019, 02:30 PM #15 Math Team     Joined: Jul 2011 From: Texas Posts: 2,978 Thanks: 1573 area of triangle ABD < area of sector ABD < area of triangle AFD $\dfrac{\sin{x}}{2} < \dfrac{x}{2} < \dfrac{\tan{x}}{2}$ $\sin{x} < x < \tan{x}$ divide every term by $\sin{x} > 0$ $1 < \dfrac{x}{\sin{x}} < \dfrac{1}{\cos{x}}$ reciprocate each term (switches the direction of the inequality) $\cos{x} < \dfrac{\sin{x}}{x} < 1$ now take the limit as $x \to 0$ ... Thanks from topsquark, babaliaris and SDK
July 9th, 2019, 04:43 AM   #16
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 Originally Posted by skeeter area of triangle ABD < area of sector ABD < area of triangle AFD $\dfrac{\sin{x}}{2} < \dfrac{x}{2} < \dfrac{\tan{x}}{2}$ $\sin{x} < x < \tan{x}$ divide every term by $\sin{x} > 0$ $1 < \dfrac{x}{\sin{x}} < \dfrac{1}{\cos{x}}$ reciprocate each term (switches the direction of the inequality) $\cos{x} < \dfrac{\sin{x}}{x} < 1$ now take the limit as $x \to 0$ ...
Nice one but I can't calculate the area of the sector. Did you use integration for that? Can you show me?

July 9th, 2019, 04:57 AM   #17
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 Originally Posted by babaliaris Nice one but I can't calculate the area of the sector. Did you use integration for that? Can you show me?
just a standard formula from geometry ... no integration necessary

$A=\dfrac{x}{2\pi} \cdot \pi r^2$

note that $r=1$

July 9th, 2019, 05:04 AM   #18
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Quote:
 Originally Posted by skeeter area of triangle ABD < area of sector ABD < area of triangle AFD $\dfrac{\sin{x}}{2} < \dfrac{x}{2} < \dfrac{\tan{x}}{2}$ $\sin{x} < x < \tan{x}$ divide every term by $\sin{x} > 0$ $1 < \dfrac{x}{\sin{x}} < \dfrac{1}{\cos{x}}$ reciprocate each term (switches the direction of the inequality) $\cos{x} < \dfrac{\sin{x}}{x} < 1$ now take the limit as $x \to 0$ ...
On an unrelated note... This seems rather difficult to formalize if you want everything rigorously. Basically, you will want a rigorous theory of area (=measure theory) first, before you can make this argument. It's intuitive of course, but I wonder what would be involved to truly make it work.

Last edited by skipjack; July 9th, 2019 at 03:12 PM.

July 9th, 2019, 02:17 PM   #19
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Quote:
 Originally Posted by babaliaris Nice one but I can't calculate the area of the sector. Did you use integration for that? Can you show me?
You don't need to compute any areas here at all. You just need to notice that the first triangle is completely contained in the sector, which is completely contained in the second triangle. The inequality follows without ever knowing any of these areas.

July 9th, 2019, 04:01 PM   #20
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Quote:
 Originally Posted by SDK You don't need to compute any areas here at all. You just need to notice that the first triangle is completely contained in the sector, which is completely contained in the second triangle. The inequality follows without ever knowing any of these areas.
I understand that, but what I don't know is how the left term of the inequality produced (sin x)/2 and how the right term produced (tan x)/2. Angle AEB is right, but isn't in those triangles.

area of triangle ABD < area of sector ABD < area of triangle AFD

Last edited by skipjack; July 12th, 2019 at 04:18 AM.

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