Welcome back to AP Calculus with Fiveable! This topic focuses on determining the of a based on information given about other functions that bound it. Weโve worked through determining limits through algebraic manipulation, graphs, and tables, so let's keep building our limit skills. ๐
โย Squeeze Theorem
Before we get into the nitty gritty, be sure to review some of the content weโve already went over!
๐ Background Knowledge
To effectively use the , you should be familiar with:
Limits: Understanding how functions behave near a specific value.
Basic Function Behavior: Knowledge about how functions like sine, cosine, exponential, etc., behave for different inputs.
๐งฉ What is the Squeeze Theorem?
The squeeze theorem states that if f(x)โคg(x)โคh(x) and limxโaโf(x)=limxโaโh(x)=L, then limxโaโg(x) must also =L. Take a look at the visual below!
We can see that the function g(x) is sandwiched between f(x) and h(x), so it must follow the same rule in the shown interval. limxโaโg(x)=L
๐งฎย Squeeze Theorem Practice Problems
Letโs work on a few questions and make sure we have the concept down!
1) Squeeze Theorem Logic
Functions g and h are twice-differentiable functions with g(2)=h(2)=4.ย It is known that g(x)<h(x) for 1<x<3. Let k be a function satisfying g(x)โคk(x)โคh(x) for 1<x<3. Is k continuous at x=2? Justify your answer.
Once youโre ready, keep on reading to see how to approach this question. โฌ๏ธ
If functions g and h are twice-differentiable, they must be continuous. Therefore, limxโ2โg(x)=4 and limxโ2โh(x)=4. Since g(x)โคk(x)โคh(x) and the conditions for continuity are met, the squeeze theorem for k(x) applies at x=2. So, limxโ2โk(x)=4 .
Since 4=g(2)<k(2)<h(2)=4, k(2) must equal 4.
We can then conclude that k(x) is continuous at x=4 because limxโ2โk(x)=k(2)=4 . Brush up on continuity rules with this guide here: Confirming Continuity Over an Interval.
This question is from the 2019 AP Calculus AB examination administered by College Board. All credit to College Board. Way to go! ๐
2) Computing a Limit Using Squeeze Theorem
Find the limit of the function g(x)=xcos(x1โ) as x 0, using the Squeeze Theorem.
In this case, we can use the fact that โ1<cos(x1โ)<1 for all x to create a bounding function.
Multiply the inequality by x, and then consider the bounding functions f(x)=โx and h(x)=x so that โx<xcos(x1โ)<x.
Since f(x)โคg(x)โคh(x) , and the functions are known to be continuous, the Squeeze Theorem can be applied. Letโs check the limits of the bounding functions as they approach 0 to see if they squeeze g(x) at x=0.
limxโ0โf(x)=limxโaโโx=0
limxโ0โh(x)=limxโaโx=0
Because limxโ0โf(x)=limxโ0โh(x)=0, the Squeeze Theorem holds true, andโฆ
xโ0limโg(x)=xโ0limโxcos(x1โ)=0
Check out the graph below to confirm our answer visually!
Graph proving limxโ0โxcos(x1โ)=0 by the Squeeze Theorem
Graph created with Desmos
You nailed it! This was a tough one. ๐ช
๐ย Closing
Great work! ๐ย The squeeze theorem is a key foundational idea for AP Calculus. You can anticipate encountering questions involving limits and the squeeze theorem on the exam, both in multiple-choice and as part of a free response.