#AP Calculus AB/BC: Squeeze Theorem - Your Ultimate Guide 🚀

Hey there, future calculus champ! 👋 Let's dive into the Squeeze Theorem, a super useful tool for finding limits. This guide is designed to make sure you're feeling confident and ready to ace those AP Calculus questions! Let's get started!

#1.8: Determining Limits Using the Squeeze Theorem

This section focuses on using the Squeeze Theorem to determine limits when direct methods don't work. It's all about bounding a tricky function between two easier ones! Remember, this concept often appears in both multiple-choice and free-response questions. Let’s break it down!

The Squeeze Theorem (also known as the Sandwich Theorem) states that if a function is always between two other functions, and those two functions approach the same limit, then the function in the middle must also approach that same limit. Think of it like being squeezed between two friends who are walking to the same spot—you’ll end up there too! 🚶‍♂️🚶‍♀️

Formally:

If $\textcolor{red}{f(x)} \leq \textcolor{blue}{ g(x)} \leq \textcolor{green}{h(x)}$ and $\lim_{{x \to a}} f(x) = \lim_{{x \to a}} h(x) = \textcolor{orange}{L}$, then $\lim_{{x \to a}} g(x) = \textcolor{orange}{L}$.

Caption: The graph shows how g(x) is 'sandwiched' between f(x) and h(x), forcing it to approach the same limit L as x approaches a.

#📚 Background Knowledge

Before we jump into practice, make sure you're comfortable with these concepts:

• Limits: How functions behave as they approach a specific value. If you need a refresher, check out this guide.
• Basic Function Behavior: Understanding the behavior of functions like sine, cosine, and exponentials. For example, $\sin(x)$ and $\cos(x)$ are always between -1 and 1. 💡

#🧮 Squeeze Theorem Practice Problems

Let's tackle some problems to solidify your understanding! Remember, practice makes perfect. 💪

#1) Squeeze Theorem Logic

Question: Functions $g$ and $h$ are twice-differentiable with $g(2) = h(2) = 4$. It's known that $g(x) < h(x)$ for $1 < x < 3$. Let $k$ be a function satisfying $g(x) \leq k(x) \leq h(x)$ for $1 < x < 3$. Is $k$ continuous at $x = 2$? Justify your answer.

Solution:

Since $g$ and $h$ are twice-differentiable, they are also continuous. Therefore,

$\lim_{{x \to 2}} g(x) = 4$ and $\lim_{{x \to 2}} h(x) = 4$.

Given that $g(x) \leq k(x) \leq h(x)$, and the limits of $g(x)$ and $h(x)$ are equal at $x=2$, the Squeeze Theorem applies to $k(x)$. Thus, $\lim_{{x \to 2}} k(x) = 4$.

Also, since $g(2) \le k(2) \le h(2)$ and $g(2) = h(2) = 4$, it follows that $k(2) = 4$.

Therefore, $k(x)$ is continuous at $x = 2$ because $\lim_{{x \to 2}} k(x) = k(2) = 4$.

#2) Computing a Limit Using Squeeze Theorem

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

Solution:

We know that $-1 \leq \cos\left(\frac{1}{x}\right) \leq 1$ for all $x$. Multiplying by $x$, we get:

$-x \leq x\cos\left(\frac{1}{x}\right) \leq x$ (for $x > 0$)

and

$x \leq x\cos\left(\frac{1}{x}\right) \leq -x$ (for $x < 0$)

Let's consider the bounding functions $\textcolor{red}{f(x) = -|x|}$ and $\textcolor{green}{h(x) = |x|}$ so that $f(x) \leq g(x) \leq h(x)$.

Now, let's find the limits of the bounding functions as $x$ approaches 0:

$\lim_{{x \to 0}} f(x) = \lim_{{x \to 0}} -|x| = 0$

$\lim_{{x \to 0}} h(x) = \lim_{{x \to 0}} |x| = 0$

Since $\lim_{{x \to 0}} f(x) = \lim_{{x \to 0}} h(x) = 0$, by the Squeeze Theorem:

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

Caption: This graph visually demonstrates how the function xcos(1/x) is squeezed between -|x| and |x|, converging to 0 as x approaches 0.

#🌟 Closing

Fantastic job! 🎉 You've now grasped the Squeeze Theorem, a powerful tool for finding limits. Remember, this theorem is particularly useful when dealing with functions that oscillate or are difficult to evaluate directly. Keep practicing, and you'll be a pro in no time!

Here’s a quick recap of the steps:

1. Identify the Function: Determine the function for which you need to find the limit.
2. Find the 'Squeeze' Functions: Locate two functions that bound the given function.
3. Ensure Known Limits: Confirm that the limits of the bounding functions are equal as x approaches the target value.

#Final Exam Focus

Okay, let's get down to the nitty-gritty of exam prep. Here's what you need to focus on for the Squeeze Theorem:

• Identifying When to Use It: Look for oscillating functions (like $\sin(1/x)$ or $\cos(1/x)$) or when direct substitution doesn't work.
• Finding Appropriate Bounding Functions: Use known inequalities (like $-1 \leq \sin(x) \leq 1$) to create the 'squeeze'.
• Justifying Your Answer: Always show that the limits of the bounding functions are equal before applying the Squeeze Theorem.
• Common Question Types: Expect to see Squeeze Theorem problems in both multiple-choice and free-response sections. They often involve trigonometric or rational functions.

#Last-Minute Tips

• Double-Check Inequalities: Make sure your inequalities are correct, especially when multiplying by a variable that could be positive or negative.
• Show All Steps: In free-response questions, clearly show each step of your reasoning. This can earn you partial credit even if you make a small mistake.
• Stay Calm: You've got this! Take deep breaths and tackle each question methodically.

#Practice Questions

Practice Question

#Multiple Choice Questions

1. If $f(x) \leq g(x) \leq h(x)$ for all $x$ near $a$, and $\lim_{x \to a} f(x) = 5$ and $\lim_{x \to a} h(x) = 5$, then $\lim_{x \to a} g(x)$ is: (A) 0 (B) 5 (C) 10 (D) Does not exist

2. What is $\lim_{x \to 0} x^2 \sin(\frac{1}{x})$? (A) 0 (B) 1 (C) Does not exist (D) $\infty$

3. Given that $-|x| \leq k(x) \leq |x|$, what is $\lim_{x \to 0} k(x)$? (A) -1 (B) 0 (C) 1 (D) Does not exist

#Free Response Question

Consider the function $f(x) = x^2\cos(\frac{1}{x})$ for $x \neq 0$.

(a) Use the Squeeze Theorem to find $\lim_{x \to 0} f(x)$.

(b) Is $f(x)$ continuous at $x=0$? Justify your answer.

(c) Find the derivative of $f(x)$.

#Scoring Guide for FRQ

(a) Use the Squeeze Theorem to find $\lim_{x \to 0} f(x)$. (3 points)

• 1 point: States that $-1 \leq \cos(\frac{1}{x}) \leq 1$

• 1 point: Multiplies by $x^2$ to get $-x^2 \leq x^2\cos(\frac{1}{x}) \leq x^2$

• 1 point: Correctly applies the Squeeze Theorem and states the limit is 0. (b) Is $f(x)$ continuous at $x=0$? Justify your answer. (2 points)

• 1 point: States that $f(0) = 0$ (or defines it to be 0 for continuity)

• 1 point: Concludes that $f(x)$ is continuous at $x=0$ because $\lim_{x \to 0} f(x) = f(0) = 0$

(c) Find the derivative of $f(x)$. (3 points)

• 1 point: Correctly applies the product rule
• 1 point: Correctly applies the chain rule
• 1 point: Correctly states the derivative $f'(x) = 2x\cos(\frac{1}{x}) + \sin(\frac{1}{x})$

Let's do this! You're going to rock this exam. 🌟