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

 Calculus Calculus Math Forum

 July 5th, 2019, 03:23 PM #1 Senior Member   Joined: Oct 2015 From: Greece Posts: 139 Thanks: 8 I can't show why x->0 of sin(x)/x is 1!!! I started using $\displaystyle -|x| \leq \sin x \leq |x|$ then $\displaystyle -\frac{|x|}{x} \leq \frac{\sin x}{x} \leq \frac{|x|}{x}$ But there is not a single value L for $\displaystyle \lim_{x \to 0} \frac{|x|}{x}=L$ . Well I have a theory but I'm not sure if it stands. $\displaystyle -\lim_{x \to 0^-}\frac{|x|}{x} = -(-1) = 1$ $\displaystyle \lim_{x \to 0^+}\frac{|x|}{x} = 1$ So $\displaystyle \lim_{x \to 0} \frac{\sin x}{x}=1$ But can I actually do that? I took different kind of limits in both cases. This seems like I'm breaking the inequality since I'm taking different limits at once! It feels like I'm doing something like this: $\displaystyle x + 5 = 5 <=> (x + 5) \cdot 10 = 5 \cdot 20$ I multiplied the two sides of the equation with not the same number. Last edited by skipjack; July 5th, 2019 at 10:50 PM. July 5th, 2019, 03:46 PM #2 Senior Member   Joined: Sep 2015 From: USA Posts: 2,641 Thanks: 1475 Thanks from topsquark and babaliaris July 5th, 2019, 07:07 PM #3 Senior Member   Joined: Oct 2018 From: USA Posts: 102 Thanks: 77 Math Focus: Algebraic Geometry The Calculus way to go about this would be L'hopitals rule: If $\lim_{x \to a} \frac{f(x)}{g(x)}$ is indeterminate ($\frac{0}{0}$, $\frac{\infty}{\infty}$), then $\displaystyle \lim_{x \to a}\frac{f(x)}{g(x)} = \lim_{x \to a}\frac{f^{\prime}(x)}{g^{\prime}(x)}$ So, since $\lim_{x \to 0} \frac{\sin{x}}{x} = \frac{0}{0}$ $\displaystyle \lim_{x \to 0} \frac{\sin{x}}{x} = \lim_{x \to 0}\frac{\cos{x}}{1} = \cos{0}=1$ But the method Romsek sent is a much more concrete way to go about it. Thanks from babaliaris July 5th, 2019, 11:03 PM   #4
Global Moderator

Joined: Dec 2006

Posts: 21,113
Thanks: 2327

Quote:
 Originally Posted by Greens The Calculus way to go about this . . .
You assume that the derivative of sin(x) is cos(x), but the usual proof of that assumes knowledge of the limit you're trying to find.

Last edited by skipjack; July 6th, 2019 at 12:44 AM. July 5th, 2019, 11:37 PM #5 Senior Member   Joined: Oct 2018 From: USA Posts: 102 Thanks: 77 Math Focus: Algebraic Geometry That's absolutely true, which is why I mentioned the other method is much more concrete. But generally, if I were tossed this problem without having to build from the ground up I would use L'hôpital. Last edited by skipjack; July 6th, 2019 at 12:43 AM. July 6th, 2019, 01:24 AM   #6
Senior Member

Joined: Oct 2015
From: Greece

Posts: 139
Thanks: 8

Quote:
 Originally Posted by Greens The Calculus way to go about this would be L'hopitals rule: If $\lim_{x \to a} \frac{f(x)}{g(x)}$ is indeterminate ($\frac{0}{0}$, $\frac{\infty}{\infty}$), then $\displaystyle \lim_{x \to a}\frac{f(x)}{g(x)} = \lim_{x \to a}\frac{f^{\prime}(x)}{g^{\prime}(x)}$ So, since $\lim_{x \to 0} \frac{\sin{x}}{x} = \frac{0}{0}$ $\displaystyle \lim_{x \to 0} \frac{\sin{x}}{x} = \lim_{x \to 0}\frac{\cos{x}}{1} = \cos{0}=1$ But the method Romsek sent is a much more concrete way to go about it.
I tried to understand Romsek's answer but I could not understand the math. Since I'm re-studying calculus, I perfectly understand your answer even currently in the book I haven't even reach the derivatives chapter (I'm still on limits and now entering continuity).

Also my original post of taking the left handed side and the right handed side limit is wrong write?
−limx→0−|x|x=−(−1)=1

limx→0+|x|x=1

So limx→0sinxx=1 July 6th, 2019, 07:34 AM #7 Senior Member   Joined: Oct 2018 From: USA Posts: 102 Thanks: 77 Math Focus: Algebraic Geometry When you took $-\lim_{x \to 0^{-}} \frac{|x|}{x}$ , the added negative sign breaks it since the fraction is already negative on the left. This creates the squeeze $\displaystyle \frac{|x|}{x} \leq \frac{\sin{x}}{x} \leq \frac{|x|}{x}$ Which would imply $\frac{|x|}{x} = \sin{x}$ , which isn't always true. Thanks from babaliaris July 6th, 2019, 02:53 PM #8 Senior Member   Joined: Sep 2016 From: USA Posts: 684 Thanks: 459 Math Focus: Dynamical systems, analytic function theory, numerics So much is wrong in this post which isn't really worth trying to correct. Two things are bad enough they should be pointed out. 1. You absolutely can NOT use L'Hospital's rule to "show" that $\lim\limits_{x \to 0} \frac{\sin x}{x} = 1$. At best this is circular logic, but I would say it's even worse than that. It is a complete misunderstanding of what L'Hospital's rule means. 2. Romsek's link is technically accurate, but I don't think it's helpful. It begins with 2 inequalities which "can be shown". But given these 2 inequalities it's completely trivial to compute this limit. The important part (which also has all of the insight here) is showing those two inequalities. This is typically done by a geometric argument which is difficult to carry out without drawing the right picture. Last edited by skipjack; July 7th, 2019 at 02:06 AM. July 6th, 2019, 03:09 PM   #9
Senior Member

Joined: Sep 2015
From: USA

Posts: 2,641
Thanks: 1475

Quote:
 Originally Posted by SDK So much is wrong in this post which isn't really worth trying to correct. Two things are bad enough they should be pointed out.
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? July 8th, 2019, 06:29 AM   #10
Senior Member

Joined: Oct 2015
From: Greece

Posts: 139
Thanks: 8

Quote:
 Originally Posted by SDK So much is wrong in this post which isn't really worth trying to correct. Two things are bad enough they should be pointed out. 1. You absolutely can NOT use L'Hospital's rule to "show" that $\lim\limits_{x \to 0} \frac{\sin x}{x} = 1$. At best this is circular logic, but I would say it's even worse than that. It is a complete misunderstanding of what L'Hospital's rule means. 2. Romsek's link is technically accurate, but I don't think it's helpful. It begins with 2 inequalities which "can be shown". But given these 2 inequalities it's completely trivial to compute this limit. The important part (which also has all of the insight here) is showing those two inequalities. This is typically done by a geometric argument which is difficult to carry out without drawing the right picture.
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 $\displaystyle \frac{0}{0}$ or $\displaystyle \frac{\infty}{\infty}$ you can use L'Hospital's rule. Tags show, sinx or x, x>0, x>0 Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post muneeb977 Linear Algebra 2 January 28th, 2017 05:28 AM fahad nasir Trigonometry 3 October 8th, 2016 05:49 PM zgonda Algebra 2 August 22nd, 2010 11:18 AM notnaeem Real Analysis 4 August 16th, 2010 01:32 PM naserellid Algebra 0 August 15th, 2010 11:25 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top      