# Use of Squeezing Theorem to Find Limits of Mathematical Functions

## Squeezing Theorem.

If f, g and h are functions such that
f(x) ≤ g(x) ≤ h(x)
for all values of x in some open interval containing a and if lim x→a f(x) = lim x→a h(x) = L then
limx→a g(x) = L
How to use the squeezing theorem? Examples are presented below.

 Example 1: Find the limit limx→0 x 2 cos(1/x) Solution to Example 1: As x approaches 0, 1 / x becomes very large in absolute value and cos(1 / x) becomes highly oscillatory. However cos(1 / x) takes values within the interval [-1,1] which is the range of cos x, hence -1 ≤ cos (1/x) ≤ 1 Multiply all terms of the above inequality by x 2 (x not equal to 0) - x 2 ≤ x 2 cos (1/x) ≤ x 2 The above inequality holds for any value of x except 0 where x 2 cos (1/x) in undefined. As x approaches 0 both - x 2 and x 2 approach 0 and according to the squeezing theorem we obtain limx→0 x 2 cos(1/x) = 0 Example 2: Find the limit limx→0 sin x / x Solution to Example 2: Assume that 0 < x < Pi/2 and let us us consider the unit circle, shown below, and a sector OAC with central angle x where x is in standard position. A is a point on the unit circle and AB is tangent to the circle at A. 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 limx→0 cos x = 1 and limx→0 1 = 1 , we can apply the squeezing theorem to obtain limx→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