How to prove lim x→infinite sin(x)/x=0?
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HallsofIvy
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Starting from what definitions?
If you are defining "sin(x)" in terms of the usual trig function (with x understood to be measured in radians) then the proof used in most Calculus text book is "geometric", using a diagram with angle x above the horizontal axis in a circle of radius 1 so that x is also the length of the arc from the end of the angle to the horizontal axis while sin(x) is the length of the vertical from that end to the horizonal axis, then show geometrically that ratio of those two lengths goes to 1.
But another perfectly valid way to do this (and one I prefer) is to define sin(x) to be the power series \(\displaystyle \sum_{n= 0}^\infty (-1)^n \frac{x^{2n+1}}{(2n+1)!}= x- \frac{x^3}{3!}+ \frac{x^5}{5!}- \frac{x^7}{7!}+ \cdot\cdot\cdot\). With that definition, immediately, \(\displaystyle \frac{sin(x)}{x}= \sum_{n=0}^\infty (-1)^n\frac{x^{2n}}{(2n+1)!}= 1- \frac{x^2}{3!}+ \frac{x^4}{5!}- \frac{x^6}{7!}+ \cdot\cdot\cdot\) and the limit is obvous.
If you are defining "sin(x)" in terms of the usual trig function (with x understood to be measured in radians) then the proof used in most Calculus text book is "geometric", using a diagram with angle x above the horizontal axis in a circle of radius 1 so that x is also the length of the arc from the end of the angle to the horizontal axis while sin(x) is the length of the vertical from that end to the horizonal axis, then show geometrically that ratio of those two lengths goes to 1.
But another perfectly valid way to do this (and one I prefer) is to define sin(x) to be the power series \(\displaystyle \sum_{n= 0}^\infty (-1)^n \frac{x^{2n+1}}{(2n+1)!}= x- \frac{x^3}{3!}+ \frac{x^5}{5!}- \frac{x^7}{7!}+ \cdot\cdot\cdot\). With that definition, immediately, \(\displaystyle \frac{sin(x)}{x}= \sum_{n=0}^\infty (-1)^n\frac{x^{2n}}{(2n+1)!}= 1- \frac{x^2}{3!}+ \frac{x^4}{5!}- \frac{x^6}{7!}+ \cdot\cdot\cdot\) and the limit is obvous.
pka
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I would suggest that
\(\displaystyle [- 1 \le \sin (x) \le 1\quad \& \quad x \to \infty] \Rightarrow \left[x > 0{\text{ and }}\dfrac{{ - 1}}{x} \le \dfrac{{\sin (x)}}{x} \le \dfrac{1}{x}\right]\)