Calculus I, Chapter 3: The Average Rate of Change and Limits
Previous Section: 3.8 Evaluating Trigonometric Limits and Other Methods
The Squeeze Theorem is useful when a function of interest, f(x), is trapped between two other functions that we name l(x) and u(x), on an interval I. That is to say:
l(x)≤f(x)≤u(x), ∀xϵI
(we choose ‘l’ for lower and ‘u’ for upper, as in l(x) is the lower function and u(x) is the upper one)
Particularly, the Squeeze Theorem is applicable if l(x) and u(x) not only trap, but also squeeze the function at a certain point. Which means that all three of these functions have one shared limit, which helps us find the limit of the middle function if the limit is unfindable otherwise.
So lim┬(x→c)〖u(x)〗=lim┬(x→c)〖l(x)〗=L, therefore lim┬(x→c)〖f(x)〗=L
Here is a visual of a function being ‘squeezed’ between two other functions:
Strategy
The strategy is to recognize the following things. Firstly, most problems that require the squeeze theorem will have a trigonometric function as a major component of the function. Each of these trigonometric functions are made of either sine or cosine, both of which have a maximum amplitude of 1. That is to say -1≤sinθ≤1 and -1≤cosθ≤1.
Second, we can “build” on to this inequality by adding, multiplying and using other operations to each side of the function. We do this in order to “create” the function we are trying to solve in the middle of the inequality. There is one problem with this: variables. When multiplying by a variable to both sides we run into an issue with positive/negative values where the inequality isn’t actually true anymore. This will be talked about in the next part of this section.’ This can be solved by multiplying the ends by the absolute value of the variable.
Third, by evaluating the limits at the same time we are able to see if the function was successfully “squeezed” and if there is a solution! Generally, this strategy works pretty well, however if it is not the case that the limit is solved it is not an indication that the limit does not exist. It only means that another method, or new upper and lower functions, may be necessary.
The Squeeze and Positivity
What does it mean for a function to be successfully “squeezed”? It is as simple as saying that both sides of the inequality must be identical. If they are at all different, then there are values between them and thus there is some ambiguity as to the solution of the limit in the middle. For example, 2.4≤Limit≤2.7 is not successful. The solution could be 2.4, 2.464, or even 2.68. Another possible issue is if the ends don’t agree with the inequality then a mistake was made. This could be something like 4≤Limit≤-2. This is clearly wrong because it implies that -2 is larger than 4.
One of the main ways in which this second issue could arise has to do with multiplying by variables. Since a variable could be positive or negative, when multiplying it is unclear whether or not to flip the inequality symbols. Take the following example:
-1≤sin(x)≤1
-x≤xsin(x)≤x
By simply multiplying by x we have caused an error. If x where a negative number, then -x≰x. This inequality is not always true. This can be difficult to understand abstractly. Here is a visual of these three graphs:
The red line, f(x)=x, is clearly lower than the blue line, f(x)=-x, on the left side of the graph. So, really we have the following inequalities:
-x≤xsin(x)≤x if x≥0 and x≤xsin(x)≤-x if x<0
However, this is horribly confusing, and we would really prefer to avoid this. A simple solution is to somehow take only the top parts of the red and blue lines for the upper function and the opposite for the lower. This can be done by using |x| instead for the ends of the inequality.
The general rule is that if a variable occurs to an odd exponent, or can change sign based on the input, to use absolute values. Otherwise, nothing is needed. For even exponents, any number to an even exponent is positive so there is no worry about having issues with the sign.
Practice with the Squeeze Theorem
Example 1: Evaluate lim┬(x→0)〖xsin(1/x〗)
Solution: Let’s break this into a few steps.
First: The limit doesn’t exist since we cannot divide by zero. Also, the limit is unsolvable using limit laws or algebra.
Second: sin(θ) is always between 1 and -1. So |sin(θ)|≤1 and |sin(1/x)|≤1.
Third: That means -1≤sin(1/x)≤1
So Finally: -|x|≤x sin(1/x)≤|x|
As soon as the function in the middle is exactly the same as our function of interest, we can evaluate the limits.
lim┬(x→0)-|x| ≤ lim┬(x→0) x sin(1/x) ≤ lim┬(x→0) |x|
0≤ lim┬(x→0) x sin(1/x) ≤ 0
And
lim┬(x→0) x sin(1/x)=0
Here’s a picture of what we just did:
Example 2: Evaluate lim┬(x→3) 〖(x-3)〗^2 e^(cos(π/(-3+x)))-4
Solution:
As always when using the Squeeze Theorem, be should begin with the trigonometric function and work our way out in order to construct the function of interest.
-1≤cos(θ)≤1
-1≤cos(π/(-3+x))≤1
e^(-1)≤e^cos(π/(-3+x)) ≤e^1
〖(x-3)〗^2 e^(-1)≤〖〖(x-3)〗^2 e〗^cos(π/(-3+x)) ≤〖(x-3)〗^2 e
〖(x-3)〗^2 e^(-1)-4≤(x-3)^2 e^cos(π/(-3+x)) -4≤(x-3)^2 e-4
lim┬(x→3) 〖(x-3)〗^2 e^(-1)-4≤lim┬(x→3) (x-3)^2 e^cos(π/(-3+x)) -4≤lim┬(x→3) (x-3)^2 e-4
0-4≤lim┬(x→3) (x-3)^2 e^cos(π/(-3+x)) -4≤0-4
lim┬(x→3) (x-3)^2 e^cos(π/(-3+x)) -4=-4
Review
The Squeeze Theorem applies to limits built from a trigonometric function
A Squeeze Theorem will always involve an inequality with three parts
When multiplying by a variable, use absolute values in cases where the sign of the output may change
If a limit is not less than and greater than the same number then it was not properly ‘squeezed’
Equations
-1≤cos(θ)≤1
-1≤sin(θ)≤1