One of the important limit when you learn calculus is limit of ratio of an angle and its sines. We got the value from this steps. Let we have a pie on a circle that have a central point O and the pie was sign on a arc AB. Let OA be horisontal line. From B we determine projection point on OA called C, that the area of triangle OBC always less than pie OAB. Next extend OB to point D that the projector of D on OA is A itself. That area of triangle OAD always greater than pie OAB ones.
|OC||BC|:2 < angle(AOB) . |OA|^2 : 2 < |OA||AD| : 2
Using a proportion of trigonometry, we know that |OC| = |OA| sin angle(AOB), |AD| = |OA| tan angle(AOB)
sin(angle(AOB)) < angle(AOB) < tan(angle(AOB))
We get the ratio of an angle and its sines by dividing all with sin(angle(AOB)), if the angle we symbolize with x, so we get,
1 < x/sin x < 1/cos x
We know that lim c = c, for a constant c, and limit 1/cos x equal to 1 for x toward 0. On advance of calculus, you will find that if a<b<c, lim a = lim c, so lim b = lim a = lim c. With it, we got
lim(x -> 0) x/sin x = 1
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