Limits of trigonometric functions

Order is kept by limits: if f(x) ≤ g(x) near a, and both limits exist, then lim f ≤ lim g.

Sandwich theorem: if f(x) ≤ g(x) ≤ h(x) near a and lim f = lim h = l, then lim g = l as well.

Comparing the areas of two triangles and a sector cut from a unit circle gives cos x < (sin x)/x < 1 for 0 < |x| < π/2.

Both cos x and 1 tend to 1 as x → 0, so the sandwich gives lim_{x→0} (sin x)/x = 1. Here x must be in radians.

It follows that lim_{x→0} (1 − cos x)/x = 0.

Worked: lim_{x→0} sin 4x/sin 2x = [4 · (sin 4x/4x)] / [2 · (sin 2x/2x)] → 4/2 = 2.

Worked: lim_{x→0} tan x/x = lim (sin x/x)·(1/cos x) = 1 · 1 = 1.

General tactic for 0/0 forms: find the factor that vanishes at the point and remove it, by factorising or by building sin x/x.