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October 11th, 2017, 10:38 AM   #1
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Math Focus: dynamical systen theory
A proof for a theorem

Dear MyMathForum Community:
Could someone please furnish me with a proof for the following?

when the limit of x --> 0, sin(x)/x = 1.

Some of you may recognize this as the "squeeze theorem" or the "pinch
theorem." Thank you.
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October 11th, 2017, 11:28 AM   #2
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you've asked this before, haven't you?

no matter ... a google search yields a nice hit

SqueezeTheorem.pdf
Thanks from Carl James Mesaros
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October 12th, 2017, 04:40 AM   #3
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No, that's not the "squeeze theorem". The "squeeze theorem" involves three functions:

If $\displaystyle f(x)\le g(x)\le h(x)$ for all x and $\displaystyle \lim_{x\to a} f(x)= \lim_{x\to a} g(x)= L$ then $\displaystyle \lim_{x\to a} g(x)= L$.

Did you mean that you can use the "squeeze theorem" to prove that $\displaystyle \lim_{x\to 0} \frac{sin(x)}{x}= 1$?
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October 12th, 2017, 06:53 AM   #4
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Math Focus: dynamical systen theory
An apology and a clarification

Dear Country Boy,
Upon further reading of the proof I desire and the "squeeze theorem", I
can say that they are two entirely different concepts. The only thing I want is
a proof for: as x --> 0, sin(x)/x = 1. Thank you.

Last edited by skipjack; October 12th, 2017 at 08:20 AM.
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October 12th, 2017, 07:05 AM   #5
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Quote:
Originally Posted by Carl James Mesaros View Post
The only thing I want is
a proof for: as x --> 0, sin(x)/x = 1
Did you check the link in my previous post? A proof of the limit in question is provided as an example of using the squeeze theorem.
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October 12th, 2017, 08:50 AM   #6
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Quote:
Originally Posted by Carl James Mesaros View Post
The only thing I want is
a proof for: as x --> 0, sin(x)/x = 1. Thank you.
Try looking at the links you've been given in this thread and the last one.
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October 12th, 2017, 10:50 AM   #7
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Math Focus: dynamical systen theory
Dear Skeeter:
I FINALLY got around to the link SqueezeTheorem.pdf. Thanks.
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October 12th, 2017, 11:52 AM   #8
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A mere three days after you saw the first responses to your first thread asking for the proof. Can you tell me why anybody should bother answering your threads?
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