# help calculus problem

##### New member
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.

#### galactus

##### Super Moderator
Staff member
$$\displaystyle \L\\0<\frac{1}{2}sin(x)<\frac{x}{2}<\frac{1}{2}tan(x)$$

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.