July 8th, 2019, 05:31 AM  #11 
Senior Member Joined: Oct 2009 Posts: 863 Thanks: 328  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  
Senior Member Joined: Oct 2015 From: Greece Posts: 137 Thanks: 8  Quote:
 
July 8th, 2019, 05:36 AM  #13 
Senior Member Joined: Oct 2009 Posts: 863 Thanks: 328 
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.

July 8th, 2019, 12:50 PM  #14  
Senior Member Joined: Sep 2016 From: USA Posts: 642 Thanks: 406 Math Focus: Dynamical systems, analytic function theory, numerics  Quote:
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: 3,016 Thanks: 1600  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 $x \to 0$ ... 
July 9th, 2019, 04:43 AM  #16  
Senior Member Joined: Oct 2015 From: Greece Posts: 137 Thanks: 8  Quote:
 
July 9th, 2019, 04:57 AM  #17 
Math Team Joined: Jul 2011 From: Texas Posts: 3,016 Thanks: 1600  
July 9th, 2019, 05:04 AM  #18  
Senior Member Joined: Oct 2009 Posts: 863 Thanks: 328  Quote:
Last edited by skipjack; July 9th, 2019 at 03:12 PM.  
July 9th, 2019, 02:17 PM  #19 
Senior Member Joined: Sep 2016 From: USA Posts: 642 Thanks: 406 Math Focus: Dynamical systems, analytic function theory, numerics  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  
Senior Member Joined: Oct 2013 From: New York, USA Posts: 661 Thanks: 87  Quote:
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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show, sinx or x, x>0, x>0 
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