help calculus problem

[ladyazpy]

In, calculus, the ratio (sin x)/x comes up naturally, and the problem is to determine what the ratio approaches when x approches 0. can you guess?
(Note that the ratio is not defined when x equals 0)

A. Referring to the figure, show the following:

A sub 1 = area of triangle QAP = (sin x) / (2)

A sub 2 = area of sector OAP = (x) / (2)

A sub 3 = area of triangle OAB = (tan x) / (2)


B. Using the fact that A sub 1 < A sub 2 < A sub 3, which we can see clearly in the figure, show that

cos x < (sin x) / (x) < 1 x > 0


C. From the inequality in part B, explain how you can conclude that for x > 0, (sin x) / x approaches 0.


please help.... this is a calculus problem in my trig. book

 

[galactus]

\(\displaystyle \L\\0<\frac{1}{2}sin(x)<\frac{x}{2}<\frac{1}{2}tan(x)\)

Multiply through by \(\displaystyle \L\\\frac{2}{sin(x)}\), then take reciprocals.

Of course, this is under the assumption that \(\displaystyle 0<x<\frac{\pi}{2}\)

The inequality also holds if \(\displaystyle \frac{-\pi}{2}<x<0\).

The Squeezing theorem tells you something about the limit of sin(x)/x wrt
the limit of cos(x) and 1.

 

[ladyazpy]

i still need explanation i can't figure it out somebody???

 

[Gene]

Maybe one of the other places you have posted this problem will help.
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Gene