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
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