Question 1131261
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*[illustration lim_sinx_over_x_proof_diagram_(crop).jpg]


Segment SC = *[tex \Large \sin(x)] by definition (note unit circle).  Altitude of triangle OSU is *[tex \Large \sin(x)] and the base of triangle OSU is 1, so the area of triangle OSU is *[tex \Large \frac{\sin(x)}{2}].


The area of a circle sector is given by *[tex \Large \frac{\theta{r^2}}{2}] where *[tex \Large \theta] is the subtended angle.  Hence, the area of sector OSU is simply *[tex \Large \frac{x}{2}]


Segment TU = *[tex \Large \tan(x)] by definition of the tangent function. (TU divided by OU, but OU measures 1).  Then the area of triangle OTU is *[tex \Large \frac{\tan(x)}{2}].


From the diagram, the area of triangle OSU is less than or equal to the area of sector OSU which in turn is less than or equal to triangle OTU.  Hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\sin(x)}{2}\ \leq\ \frac{x}{2}\ \leq\ \frac{\tan(x)}{2}]


Multiply by *[tex \Large \frac{2}{\sin(x)}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 1\ \leq\ \frac{x}{\sin(x)}\ \leq\ \frac{1}{\cos(x)}]


Take the reciprocal and reverse the sense of the inequalities:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos(x)\ \leq\ \frac{\sin(x)}{x}\ \leq\ 1]


But *[tex \Large \lim_{x\right{0}}\,\cos(x)\ =\ 1] and *[tex \Large \lim_{x\right{0}}\,1\ =\ 1]


Therefore, by the Squeeze Theorem:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \lim_{x\right{0}}\,\frac{\sin(x)}{x}\ =\ 1]
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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