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1.8 Determining Limits Using the Squeeze Theorem

8 min readfebruary 15, 2024

Welcome back to AP Calculus with Fiveable! This topic focuses on determining the limit of a function based on information given about other functions that bound it. We’ve worked through determining limits through algebraic manipulation, graphs, and tables, so let's keep building our limit skills. 🙌

⌛ Squeeze Theorem

Before we get into the nitty gritty, be sure to review some of the content we’ve already went over!

📚 Background Knowledge

To effectively use the Squeeze Theorem, 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.

🧩 What is the Squeeze Theorem?

The squeeze theorem states that if f(x)g(x)h(x)\textcolor{red}{f(x)} \leq \textcolor{blue}{ g(x)} \leq \textcolor{green}{h(x)} and limxaf(x)=limxah(x)=L\lim_{{x \to a}} f(x) = \lim_{{x \to a}} h(x) = \textcolor{orange}{L}, then limxag(x)\lim_{{x \to a}} g(x) must also =L= \textcolor{orange}{L}. Take a look at the visual below!

Screen Shot 2023-12-12 at 8.51.53 PM.png

Graph drawn using Virtual Graph Paper

We can see that the function g(x)\textcolor {blue}{g(x)} is sandwiched between f(x)\textcolor {red}{f(x)} and h(x)\textcolor {green}{h(x)}, so it must follow the same rule in the shown interval. limxag(x)=L\lim_{{x \to a}} g(x) = \textcolor{orange}{L}


🧮 Squeeze Theorem Practice Problems

Let’s work on a few questions and make sure we have the concept down!

1) Squeeze Theorem Logic

Functions gg and hh are twice-differentiable functions with g(2)=h(2)=4.g(2) = h(2) = 4. It is known that g(x)<h(x)g(x) < h(x) for 1<x<31 < x < 3. Let k be a function satisfying g(x)k(x)h(x)g(x) \leq k(x) \leq h(x) for
1<x<31 < x < 3. Is kk continuous at x=2x = 2? Justify your answer.

Once you’re ready, keep on reading to see how to approach this question. ⬇️

If functions gg and hh are twice-differentiable, they must be continuous. Therefore, limx2g(x)=4\lim_{{x \to 2}} g(x) = 4 and limx2h(x)=4\lim_{{x \to 2}} h(x) = 4. Since g(x)k(x)h(x)g(x) \leq k(x) \leq h(x) and the conditions for continuity are met, the squeeze theorem for k(x)k(x) applies at x=2x = 2. So, limx2k(x)=4\lim_{{x \to 2}} k(x) = 4 .

Since 4=g(2)<k(2)<h(2)=44 = g(2) < k(2) < h(2) = 4, k(2)k(2) must equal 44.

We can then conclude that k(x)k(x) is continuous at x=4x = 4 because limx2k(x)=k(2)=4\lim_{{x \to 2}} k(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! 👏

2) Computing a Limit Using Squeeze Theorem

Find the limit of the function g(x)=xcos(1x)\textcolor{blue}{g(x) = x\cos\left(\frac{1}{x}\right)} as xx approaches 0, using the Squeeze Theorem.

In this case, we can use the fact that 1<cos(1x)<1-1 < \cos\left(\frac{1}{x}\right) < 1 for all xx to create a bounding function.

Multiply the inequality by xx, and then consider the bounding functions f(x)=x\textcolor{red}{f(x) = -x} and h(x)=x\textcolor{green}{h(x) = x} so that x<xcos(1x)<x-x < x\cos\left(\frac{1}{x}\right) < x.

Since f(x)g(x)h(x)f(x) \leq g(x) \leq h(x) , and the functions are known to be continuous, the Squeeze Theorem can be applied. Let’s check the limits of the bounding functions as they approach 0 to see if they squeeze g(x)g(x) at x=0x = 0.

limx0f(x)=limxax=0\lim_{{x \to 0}} f(x) = \lim_{{x \to a}} -x = 0

limx0h(x)=limxax=0\lim_{{x \to 0}} h(x) = \lim_{{x \to a}} x = 0

Because limx0f(x)=limx0h(x)=0\lim_{{x \to 0}} f(x) = \lim_{{x \to 0}} h(x) = 0, the Squeeze Theorem holds true, and…

limx0g(x)=limx0xcos(1x)=0\lim_{{x \to 0}} g(x) = \lim_{{x \to 0}} x\cos\left(\frac{1}{x}\right) = 0

Check out the graph below to confirm our answer visually!

Screen Shot 2023-12-12 at 10.33.48 PM.png

Graph created with Desmos

You nailed it! This was a tough one. 💪


🌟 Closing

Great work! 🙌 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.

https://media2.giphy.com/media/ur5T6Wuw4xK2afXVmd/giphy.gif?cid=7941fdc64i8t3xwgu78ov7fo1wvrrynfxrd4d9loorgxbu2d&ep=v1_gifs_search&rid=giphy.gif&ct=g

Image Courtesy of Giphy

If you’d like some steps to follow, here they are:

  1. 🤔 Identifying the Function: Recognize the function for which you need to determine the limit.
  2. 👀 Finding the 'Squeeze' Functions: Locate two functions that 'squeeze' the given function between them.
  3. 🏁 Ensuring Known Limits: Confirm that the limits of the 'squeezing' functions are known as x approaches the same value.

You got this! 🤩

Key Terms to Review (6)

Approaches

: In calculus, "approaches" refers to the behavior of a function as the input values get closer and closer to a certain value or infinity. It describes how the output values of the function get arbitrarily close to a specific number or approach a particular limit.

Bounded

: A function is bounded if its values are limited or restricted within a certain range. It means that the function does not go to infinity or negative infinity.

Function

: A relationship between two sets where each input (domain) value corresponds to exactly one output (range) value.

Limit

: The limit of a function is the value that the function approaches as the input approaches a certain value or infinity. It represents the behavior of the function near a specific point.

Oscillate

: In calculus, "oscillate" refers to the behavior of a function when it repeatedly alternates between two values without approaching any particular number. It describes functions that exhibit periodic fluctuations around one or more equilibrium points.

Squeeze Theorem

: The Squeeze Theorem states that if two functions, g(x) and h(x), both approach the same limit L as x approaches a certain value c, and another function f(x) is always between g(x) and h(x) near c (except possibly at c itself), then f(x) also approaches L as x approaches c.

1.8 Determining Limits Using the Squeeze Theorem

8 min readfebruary 15, 2024

Welcome back to AP Calculus with Fiveable! This topic focuses on determining the limit of a function based on information given about other functions that bound it. We’ve worked through determining limits through algebraic manipulation, graphs, and tables, so let's keep building our limit skills. 🙌

⌛ Squeeze Theorem

Before we get into the nitty gritty, be sure to review some of the content we’ve already went over!

📚 Background Knowledge

To effectively use the Squeeze Theorem, 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.

🧩 What is the Squeeze Theorem?

The squeeze theorem states that if f(x)g(x)h(x)\textcolor{red}{f(x)} \leq \textcolor{blue}{ g(x)} \leq \textcolor{green}{h(x)} and limxaf(x)=limxah(x)=L\lim_{{x \to a}} f(x) = \lim_{{x \to a}} h(x) = \textcolor{orange}{L}, then limxag(x)\lim_{{x \to a}} g(x) must also =L= \textcolor{orange}{L}. Take a look at the visual below!

Screen Shot 2023-12-12 at 8.51.53 PM.png

Graph drawn using Virtual Graph Paper

We can see that the function g(x)\textcolor {blue}{g(x)} is sandwiched between f(x)\textcolor {red}{f(x)} and h(x)\textcolor {green}{h(x)}, so it must follow the same rule in the shown interval. limxag(x)=L\lim_{{x \to a}} g(x) = \textcolor{orange}{L}


🧮 Squeeze Theorem Practice Problems

Let’s work on a few questions and make sure we have the concept down!

1) Squeeze Theorem Logic

Functions gg and hh are twice-differentiable functions with g(2)=h(2)=4.g(2) = h(2) = 4. It is known that g(x)<h(x)g(x) < h(x) for 1<x<31 < x < 3. Let k be a function satisfying g(x)k(x)h(x)g(x) \leq k(x) \leq h(x) for
1<x<31 < x < 3. Is kk continuous at x=2x = 2? Justify your answer.

Once you’re ready, keep on reading to see how to approach this question. ⬇️

If functions gg and hh are twice-differentiable, they must be continuous. Therefore, limx2g(x)=4\lim_{{x \to 2}} g(x) = 4 and limx2h(x)=4\lim_{{x \to 2}} h(x) = 4. Since g(x)k(x)h(x)g(x) \leq k(x) \leq h(x) and the conditions for continuity are met, the squeeze theorem for k(x)k(x) applies at x=2x = 2. So, limx2k(x)=4\lim_{{x \to 2}} k(x) = 4 .

Since 4=g(2)<k(2)<h(2)=44 = g(2) < k(2) < h(2) = 4, k(2)k(2) must equal 44.

We can then conclude that k(x)k(x) is continuous at x=4x = 4 because limx2k(x)=k(2)=4\lim_{{x \to 2}} k(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! 👏

2) Computing a Limit Using Squeeze Theorem

Find the limit of the function g(x)=xcos(1x)\textcolor{blue}{g(x) = x\cos\left(\frac{1}{x}\right)} as xx approaches 0, using the Squeeze Theorem.

In this case, we can use the fact that 1<cos(1x)<1-1 < \cos\left(\frac{1}{x}\right) < 1 for all xx to create a bounding function.

Multiply the inequality by xx, and then consider the bounding functions f(x)=x\textcolor{red}{f(x) = -x} and h(x)=x\textcolor{green}{h(x) = x} so that x<xcos(1x)<x-x < x\cos\left(\frac{1}{x}\right) < x.

Since f(x)g(x)h(x)f(x) \leq g(x) \leq h(x) , and the functions are known to be continuous, the Squeeze Theorem can be applied. Let’s check the limits of the bounding functions as they approach 0 to see if they squeeze g(x)g(x) at x=0x = 0.

limx0f(x)=limxax=0\lim_{{x \to 0}} f(x) = \lim_{{x \to a}} -x = 0

limx0h(x)=limxax=0\lim_{{x \to 0}} h(x) = \lim_{{x \to a}} x = 0

Because limx0f(x)=limx0h(x)=0\lim_{{x \to 0}} f(x) = \lim_{{x \to 0}} h(x) = 0, the Squeeze Theorem holds true, and…

limx0g(x)=limx0xcos(1x)=0\lim_{{x \to 0}} g(x) = \lim_{{x \to 0}} x\cos\left(\frac{1}{x}\right) = 0

Check out the graph below to confirm our answer visually!

Screen Shot 2023-12-12 at 10.33.48 PM.png

Graph created with Desmos

You nailed it! This was a tough one. 💪


🌟 Closing

Great work! 🙌 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.

https://media2.giphy.com/media/ur5T6Wuw4xK2afXVmd/giphy.gif?cid=7941fdc64i8t3xwgu78ov7fo1wvrrynfxrd4d9loorgxbu2d&ep=v1_gifs_search&rid=giphy.gif&ct=g

Image Courtesy of Giphy

If you’d like some steps to follow, here they are:

  1. 🤔 Identifying the Function: Recognize the function for which you need to determine the limit.
  2. 👀 Finding the 'Squeeze' Functions: Locate two functions that 'squeeze' the given function between them.
  3. 🏁 Ensuring Known Limits: Confirm that the limits of the 'squeezing' functions are known as x approaches the same value.

You got this! 🤩

Key Terms to Review (6)

Approaches

: In calculus, "approaches" refers to the behavior of a function as the input values get closer and closer to a certain value or infinity. It describes how the output values of the function get arbitrarily close to a specific number or approach a particular limit.

Bounded

: A function is bounded if its values are limited or restricted within a certain range. It means that the function does not go to infinity or negative infinity.

Function

: A relationship between two sets where each input (domain) value corresponds to exactly one output (range) value.

Limit

: The limit of a function is the value that the function approaches as the input approaches a certain value or infinity. It represents the behavior of the function near a specific point.

Oscillate

: In calculus, "oscillate" refers to the behavior of a function when it repeatedly alternates between two values without approaching any particular number. It describes functions that exhibit periodic fluctuations around one or more equilibrium points.

Squeeze Theorem

: The Squeeze Theorem states that if two functions, g(x) and h(x), both approach the same limit L as x approaches a certain value c, and another function f(x) is always between g(x) and h(x) near c (except possibly at c itself), then f(x) also approaches L as x approaches c.


© 2024 Fiveable Inc. All rights reserved.

AP® and SAT® are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website.


© 2024 Fiveable Inc. All rights reserved.

AP® and SAT® are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website.