My Math Forum Concerning a limit involving sin(x)

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 October 9th, 2017, 06:01 AM #1 Member   Joined: May 2014 From: Allentown PA USA Posts: 97 Thanks: 6 Math Focus: dynamical systen theory Concerning a limit involving sin(x) Dear MyMathForum Community: According to the textbook I am using, the limit as x --> 0 of sin(x)/x =1. What would be an acceptable proof for this? Thank you.
 October 9th, 2017, 06:13 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 6,970 Thanks: 2290 Math Focus: Mainly analysis and algebra The only proofs I know of are geometric and use the squeeze theorem. For example: Signs of Limits. But it does depend on your definition of $\sin{(x)}$. If your definition is the power series, the result is straightforward (once you've prove that the series for $\cos{(x)}$ is convergent for small $x$). But I think that definition is totally unnatural. Why would you start looking at that series if you didn't know what it represented already? Last edited by v8archie; October 9th, 2017 at 06:20 AM.
October 9th, 2017, 08:07 AM   #3
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 Originally Posted by Carl James Mesaros Dear MyMathForum Community: According to the textbook I am using, the limit as x --> 0 of sin(x)/x =1. What would be an acceptable proof for this? Thank you.
This is not a rigorous proof, but using L'Hopital's rule it is straightforward to obtain the limit:

$\displaystyle \lim_{x \rightarrow 0} \frac{\sin x}{x} = \lim_{x \rightarrow 0} \frac{\cos x}{1} = 1$

I'm a lazy physicist so that's enough for me!

As for a more robust, rigorous proof... try V8Archie's suggestion

 October 9th, 2017, 09:09 AM #4 Math Team   Joined: Dec 2013 From: Colombia Posts: 6,970 Thanks: 2290 Math Focus: Mainly analysis and algebra It's not a proof at all, because you can't find the derivative of $\sin{(x)}$ without knowing the limit $\displaystyle \lim_{h \to 0} \frac{\sin{(h)}}{h}$. You might call it a demonstration. Thanks from Benit13, Carl James Mesaros and SDK
 October 9th, 2017, 02:45 PM #5 Senior Member   Joined: Sep 2016 From: USA Posts: 165 Thanks: 72 Math Focus: Dynamical systems, analytic function theory, numerics Archie is 100% correct and this illustrates why it is a huge mistake (in my opinion) to teach L'hospital's so early in calculus. This can be subtle the first time you see it, but Archie's point is the following. The limit $\lim_{x \to 0} \frac{\sin(x)}{x}$ is nothing more than the derivative of $\sin(x)$ evaluated at 0. Applying L'hospital's in this case is circular since evaluating this derivative assumes you already know the limit you are trying to compute. In general, L'hospital's rule can not be used to compute derivatives. To see why, assume for simplicity that $f \in C^1$, then applying L'hospital's rule to the derivative gives $\lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \frac{\partial}{\partial h} f(x+h) |_{h = 0} = f'(x+h)|_{h = 0} = f'(x).$ In other words, you can't compute a derivative using L'hospital's rule. Either you already know the derivative in which case you have no need to apply L'hospital's rule, or you don't in which case L'hospital's rule only tells you that the derivative is equal to the derivative, whatever that may be. In general, a limit with a linear term in the denominator is almost always the derivative of something "in disguise" which makes this rule far less useful than it seems at first glance.
 October 9th, 2017, 05:08 PM #6 Math Team   Joined: Jan 2015 From: Alabama Posts: 2,719 Thanks: 699 How you prove $\displaystyle \lim_{x\to 0} \frac{sin(x)}{x}= 1$ depends strongly upon how you define "sin(x)"! One perfectly plausible way to do that is to define $\displaystyle sin(x)= \sum_{n= 0}^\infty \frac{x^{2n+1}}{(2n+1)!}$. In that case $\displaystyle \frac{sin(x)}{x}= \sum_{n=0}^\infty \frac{x^{2n}}{(2n+1)!}= 1+ \frac{x^2}{2!}+ \cdot\cdot\cdot$. That clearly has limit 1 as x goes to 0. Another way to define sin(x) is "y= sin(x) is the function, y, satisfying y''+ y= 0 with initial values y(0)= 0, y'(0)= 1" and define cos(x) as "y= cos(x) is the function, y, satisfying y''+ y= 0 with initial values y(0)= 1, y'(0)= 0. Then we can show, without using $\displaystyle \lim_{x\to 0}\frac{sin(x)}{x}$ that sin'(x)= cos(x). Thanks from Maschke and topsquark
October 9th, 2017, 06:26 PM   #7
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 Originally Posted by Country Boy One perfectly plausible way to do that is to define $\displaystyle sin(x)= \sum_{n= 0}^\infty \frac{x^{2n+1}}{(2n+1)!}$.
I addressed this one above. In particular, I maintain that there is no reason why you would define $\sin{(x)}$ in that way unless you happen to know that the series represents $\sin{(x)}$ and you just wish to avoid certain difficulties such as how to prove the limit in question.

 October 12th, 2017, 04:57 PM #8 Math Team   Joined: Jan 2015 From: Alabama Posts: 2,719 Thanks: 699 Then you are asserting that "sin(x)" exists independently of its definition? Thanks from Joppy
 October 12th, 2017, 05:24 PM #9 Math Team   Joined: Dec 2013 From: Colombia Posts: 6,970 Thanks: 2290 Math Focus: Mainly analysis and algebra What? There are at least three mathematical definitions of $\sin{(x)}$. The function (obviously) exists independent of the particular one you pick. I'm asserting that selecting a definition that runs contrary to the development of the theory simply to avoid particular difficulties is in the spirit of the question. I would suggest that such an answer would require a detailed demonstration that the series is actually the $\sin{(x)}$ as defined in "more natural" ways. Nobody ever wondered whether the series $\displaystyle \sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)!}x^{2n+1}$ might be useful without first knowing that it was the series expansion of $\sin{(x)}$. Last edited by v8archie; October 12th, 2017 at 05:39 PM.
October 12th, 2017, 05:31 PM   #10
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 Originally Posted by Country Boy Then you are asserting that "sin(x)" exists independently of its definition?
What we call "sin(x)" has been around for thousands of years. The definition given is relatively new.

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