## 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

lim_{x→a} g(x) = L
How to use the squeezing theorem?

__Example 1:__ Find the limit lim_{x→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

lim_{x→0} x^{ 2} cos(1/x) = 0

__Example 2:__ Find the limit lim_{x→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

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

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