# Use of Squeezing Theorem to Find Limits of Mathematical Functions

## Squeezing Theorem.

If f, g and h are functions such that_{x→a}f(x) = lim

_{x→a}h(x) = L then

_{x→a}g(x) = L

Point C is a point on the unit circle (radius = 1)and has coordinates (cos x, sin x). Let us find the areas of triangle OAC, sector OAC and triangle area of triangle OAC = (1/2)*(base)*(height) = (1/2)*(1)*(y coordinate of point C) = (1/2)(sin x) Note: we have used base = radius = 1 area of sector OAC = (1/2)*(x)*(radius) ^{ 2}= (1/2) (1) ^{ 2} x = (1/2) x
area of triangle OAB = (1/2)*(base)*(height) = (1/2)*(1)*(tan x) = (1/2) tan x Comparing the three areas, we can write the inequality area of triangle OAC < area of sector OAC < area of triangle OAB Substitute the areas in the above inequality by their expressions obtained above. (1/2)(sin x) < (1/2) x < (1/2) tan x Multiply all terms by 2 / sin x gives 1 < x / sin x < 1 / cos x Take the reciprocal and reverse the two inequality symbols in the double inequality 1 > sin x / x > cos x Which the same as cos x < sin x / x < 1 It can be shown that the above inequality hols for -Pi/ 2 < x < 0 so the the above inequality hold for all x except x = 0 where sin x / x is undefined. Since lim _{x→0} cos x = 1
and lim _{x→0} 1 = 1 ,
we can apply the squeezing theorem to obtain lim _{x→0} sin x / x = 1
This result is very important and will be used to find other limits of trigonometric functions and derivatives More on limits Calculus Tutorials and Problems |