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1.8 Determining Limits Using the Squeeze Theorem

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. Weโ€™ve worked through determining limits through algebraic manipulation, graphs, and tables, so let's keep building our limit skills. ๐Ÿ™Œ

โŒ›ย Squeeze Theorem

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

๐Ÿ“š 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.

๐Ÿงฉ What is the Squeeze Theorem?

The squeeze theorem states that if f(x)โ‰คg(x)โ‰คh(x)\textcolor{red}{f(x)} \leq \textcolor{blue}{ g(x)} \leq \textcolor{green}{h(x)} and limโกxโ†’af(x)=limโกxโ†’ah(x)=L\lim_{{x \to a}} f(x) = \lim_{{x \to a}} h(x) = \textcolor{orange}{L}, then limโกxโ†’ag(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 representing squeeze theorem

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. limโกxโ†’ag(x)=L\lim_{{x \to a}} g(x) = \textcolor{orange}{L}


๐Ÿงฎย Squeeze Theorem Practice Problems

Letโ€™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.ย 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)โ‰คk(x)โ‰คh(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 youโ€™re ready, keep on reading to see how to approach this question. โฌ‡๏ธ

If functions gg and hh are twice-differentiable, they must be continuous. Therefore, limโกxโ†’2g(x)=4\lim_{{x \to 2}} g(x) = 4 and limโกxโ†’2h(x)=4\lim_{{x \to 2}} h(x) = 4. Since g(x)โ‰คk(x)โ‰คh(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, limโกxโ†’2k(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 limโกxโ†’2k(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! ๐Ÿ‘

2) Computing a Limit Using Squeeze Theorem

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

In this case, we can use the fact that โˆ’1<cosโก(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)=โˆ’x\textcolor{red}{f(x) = -x} and h(x)=x\textcolor{green}{h(x) = x} so that โˆ’x<xcosโก(1x)<x-x < x\cos\left(\frac{1}{x}\right) < x.

Since f(x)โ‰คg(x)โ‰คh(x)f(x) \leq g(x) \leq h(x) , and the functions are known to be continuous, the Squeeze Theorem can be applied. Letโ€™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.

limโกxโ†’0f(x)=limโกxโ†’aโˆ’x=0\lim_{{x \to 0}} f(x) = \lim_{{x \to a}} -x = 0

limโกxโ†’0h(x)=limโกxโ†’ax=0\lim_{{x \to 0}} h(x) = \lim_{{x \to a}} x = 0

Because limโกxโ†’0f(x)=limโกxโ†’0h(x)=0\lim_{{x \to 0}} f(x) = \lim_{{x \to 0}} h(x) = 0, the Squeeze Theorem holds true, andโ€ฆ

limโกxโ†’0g(x)=limโกxโ†’0xcosโก(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 proving limโกxโ†’0xcosโก(1x)=0\lim_{{x \to 0}}x\cos\left(\frac{1}{x}\right) = 0 by the Squeeze Theorem

Graph created with Desmos

You nailed it! This was a tough one. ๐Ÿ’ช


๐ŸŒŸย Closing

Great work! ๐Ÿ™Œย 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

Encouraging GIF with animated ice cream

Image Courtesy of Giphy

If youโ€™d like some steps to follow, here they are:

  1. ๐Ÿค” Identifying the Function: Recognize the function for which you need to determine the limit.
  2. ๐Ÿ‘€ Finding the 'Squeeze' Functions: Locate two functions that 'squeeze' the given function between them.
  3. ๐Ÿ Ensuring Known Limits: Confirm that the limits of the 'squeezing' functions are known as x approaches the same value.

You got this! ๐Ÿคฉ

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APยฎ and SATยฎ are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website.


ยฉ 2024 Fiveable Inc. All rights reserved.
APยฎ and SATยฎ are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website.

ยฉ 2024 Fiveable Inc. All rights reserved.
APยฎ and SATยฎ are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website.
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