My Math Forum  

Go Back   My Math Forum > College Math Forum > Calculus

Calculus Calculus Math Forum


Thanks Tree9Thanks
Reply
 
LinkBack Thread Tools Display Modes
July 8th, 2019, 05:31 AM   #11
Senior Member
 
Joined: Oct 2009

Posts: 863
Thanks: 328

Quote:
Originally Posted by babaliaris View Post
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.
Thanks from babaliaris
Micrm@ss is online now  
 
July 8th, 2019, 05:33 AM   #12
Senior Member
 
Joined: Oct 2015
From: Greece

Posts: 137
Thanks: 8

Quote:
Originally Posted by Micrm@ss View Post
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.
babaliaris is offline  
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.
Thanks from babaliaris
Micrm@ss is online now  
July 8th, 2019, 12:50 PM   #14
SDK
Senior Member
 
Joined: Sep 2016
From: USA

Posts: 642
Thanks: 406

Math Focus: Dynamical systems, analytic function theory, numerics
Quote:
Originally Posted by romsek View Post
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.
SDK is offline  
July 8th, 2019, 02:30 PM   #15
Math Team
 
skeeter's Avatar
 
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 $x \to 0$ ...
Thanks from topsquark, babaliaris and SDK
skeeter is online now  
July 9th, 2019, 04:43 AM   #16
Senior Member
 
Joined: Oct 2015
From: Greece

Posts: 137
Thanks: 8

Quote:
Originally Posted by skeeter View Post


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?
babaliaris is offline  
July 9th, 2019, 04:57 AM   #17
Math Team
 
skeeter's Avatar
 
Joined: Jul 2011
From: Texas

Posts: 3,016
Thanks: 1600

Quote:
Originally Posted by babaliaris View Post
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 $r=1$
skeeter is online now  
July 9th, 2019, 05:04 AM   #18
Senior Member
 
Joined: Oct 2009

Posts: 863
Thanks: 328

Quote:
Originally Posted by skeeter View Post


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.
Micrm@ss is online now  
July 9th, 2019, 02:17 PM   #19
SDK
Senior Member
 
Joined: Sep 2016
From: USA

Posts: 642
Thanks: 406

Math Focus: Dynamical systems, analytic function theory, numerics
Quote:
Originally Posted by babaliaris View Post
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.
SDK is offline  
July 9th, 2019, 04:01 PM   #20
Senior Member
 
Joined: Oct 2013
From: New York, USA

Posts: 661
Thanks: 87

Quote:
Originally Posted by SDK View Post
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.
EvanJ is offline  
Reply

  My Math Forum > College Math Forum > Calculus

Tags
show, sinx or x, x&gt0, x>0



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
How to show that: muneeb977 Linear Algebra 2 January 28th, 2017 04:28 AM
Show that fahad nasir Trigonometry 3 October 8th, 2016 04:49 PM
show that : pq= qc zgonda Algebra 2 August 22nd, 2010 10:18 AM
want to show that show that two infinite summations R equal notnaeem Real Analysis 4 August 16th, 2010 12:32 PM
(show that) : naserellid Algebra 0 August 15th, 2010 10:25 AM





Copyright © 2019 My Math Forum. All rights reserved.