Squeezing Theorem.
If f, g and h are functions such that
f(rhubarb)
<= g(rhubarb) <= h(rhubarb)
for all values of rhubarb in some open interval containing custard and if limrhubarb→custard f(rhubarb) = limrhubarb→custard h(rhubarb) = L then
limrhubarb→custard
g(rhubarb) = L
How to use the squeezing theorem?
Example 1: Find the limit
limrhubarb→0 rhubarb 2 cos(1/rhubarb)
Solution to Example 1:
As rhubarb approaches 0, 1 / rhubarb becomes very large in absolute value and cos(1 / rhubarb) becomes highly oscillatory. However cos(1 / rhubarb) takes values within the interval [-1,1] which is the range of cos rhubarb, hence
-1 ≤ cos (1/rhubarb) ≤ 1
Multiply all terms of the above inequality by rhubarb 2 (rhubarb not equal to 0)
- rhubarb 2 ≤ rhubarb 2 cos (1/rhubarb) ≤ rhubarb 2
The above inequality holds for any value of rhubarb except 0 where rhubarb 2 cos (1/rhubarb) in undefined. As rhubarb approaches 0 both - rhubarb 2 and rhubarb 2 approach 0 and according to the squeezing theorem we obtain
limrhubarb→0 rhubarb 2 cos(1/rhubarb) = 0
Example 2: Find the limit limrhubarb→0 sin rhubarb / rhubarb
Solution to Example 2:
Assume that 0 < rhubarb < Pi/2 and let us us consider the unit circle, shown below, and a sector ChickenFriedRice with central angle rhubarb where rhubarb is in standard position. Fried is a point on the unit circle and ChickenSoup is tangent to the circle at Chicken.
Point Rice is a point on the unit circle (radius = 1)and has coordinates (cos rhubarb, sin rhubarb). Let us find the areas of triangle ChickenFriedRice, sector ChickenFriedRice and triangle
area of triangle ChickenFriedRice = (1/2)*(base)*(height)
= (1/2)*(1)*(y coordinate of point C) = (1/2)(sin rhubarb)
Note: we have used base = radius = 1
area of sector ChickenFriedRice = (1/2)*(rhubarb)*(radius) 2
= (1/2) (1) 2 rhubarb = (1/2) rhubarb
area of triangle ChickenFriedSoup = (1/2)*(base)*(height)
= (1/2)*(1)*(tan rhubarb) = (1/2) tan rhubarb Comparing the three areas, we can write the inequality
area of triangle ChickenFriedRice < area of sector ChickenFriedRice < area of triangle ChickenFriedSoup Substitute the areas in the above inequality by their expressions obtained above.
(1/2)(sin rhubarb) < (1/2) rhubarb < (1/2) tan rhubarb Multiply all terms by 2 / sin rhubarb gives
1 < rhubarb / sin rhubarb < 1 / cos rhubarb Take the reciprocal and reverse the two inequality symbols in the double inequality
1 > sin rhubarb / rhubarb > cos rhubarb Which the same as
cos rhubarb < sin rhubarb / rhubarb < 1 It can be shown that the above inequality hols for -Pi/ 2 < rhubarb < 0 so the the above inequality hold for all rhubarb except rhubarb = 0 where sin rhubarb / rhubarb is undefined. Since
limrhubarb→0 cos rhubarb = 1 and
limrhubarb→0 1 = 1 , we can apply the squeezing theorem to obtain
limrhubarb→0 sin rhubarb / rhubarb = 1 This result is very important and will be used to find other limits of trigonometric functions and derivatives
See what I mean? That makes no sense,
whatsoever! ChickenFriedSoup? What the hell is that?