Determining Limits Using the Squeeze Theorem - AP Calculus AB Study Guide

The Core Idea: Determining Limits Using the Squeeze Theorem

The Squeeze Theorem provides a powerful method for finding the limit of a function that is difficult to evaluate directly, often because it involves oscillation or an indeterminate form. The core idea is to "trap" or "squeeze" the target function, $f(x)$, between two other functions, $g(x)$ and $h(x)$, whose limits are known and, crucially, are equal at the point of interest. If the outer functions both converge to the same value $L$ as $x$ approaches $c$, then the function trapped between them has no choice but to also converge to that same limit $L$. This theorem is an indirect but rigorous way to determine a limit by using the behavior of surrounding functions.

The Squeeze Theorem

The Squeeze Theorem is a formal rule for determining limits. It is stated as follows:

Let $f(x)$, $g(x)$, and $h(x)$ be functions.

If $g(x)\le f(x)\le h(x)$ for all $x$ in some open interval containing $c$, except possibly at $x=c$ itself,

and if ${\lim }_{x\to c}g(x)={\lim }_{x\to c}h(x)=L$,

then it must be true that ${\lim }_{x\to c}f(x)=L$.

Understanding the Conditions

The Squeeze Theorem is only valid when its specific conditions are met. Understanding these conditions is critical for its correct application.

• The Inequality Condition: The relationship $g(x)\le f(x)\le h(x)$ must be established and hold true for an interval of $x$-values around $c$. The theorem does not apply if this inequality is not true. This is often the most challenging part of the problem—finding or proving this bounding relationship. For many problems in AP Calculus, this involves using the known range of trigonometric functions like $\sin (x)$ or $\cos (x)$.

• The Equal Limits Condition: The most crucial part of the theorem is that the limits of the two outer (bounding) functions must be equal. If ${\lim }_{x\to c}g(x)=L$ but ${\lim }_{x\to c}h(x)=M$ where $L\ne M$, the Squeeze Theorem tells us nothing about the limit of $f(x)$. The "squeeze" only works if the top and bottom functions meet at the same point.

• Behavior at $c$ is Irrelevant: As with all limits, the value of the functions at$x=c$ does not matter. The inequality does not need to hold at $x=c$, and the functions do not even need to be defined at $x=c$. The theorem is concerned only with the behavior of the functions as $x$approaches$c$.

Core Concepts & Rules

• The Squeeze Theorem is an analytical tool used to find the limit of a function, $f(x)$, by comparing it to two other functions, $g(x)$ and $h(x)$, that bound it from below and above.

• To apply the theorem, you must first establish an inequality of the form $g(x)\le f(x)\le h(x)$ that is valid in an interval around the point $c$.

• You must then show that the limits of the outer functions, $g(x)$ and $h(x)$, both exist and are equal to the same value, $L$, as $x$ approaches $c$.

• If both the inequality condition and the equal limits condition are met, you can conclude that the limit of the inner function, $f(x)$, as $x$ approaches $c$ is also $L$.

Step-by-Step Example 1: A Classic Trigonometric Limit

Problem: Find the limit ${\lim }_{x\to 0}{x}^2\cos \left(\frac{1}{x}\right)$.

Solution:

Direct substitution results in $(0{)}^2\cos (1/0)$, which is undefined. We cannot simplify this algebraically. This structure is a classic indicator for using the Squeeze Theorem.

Step 1: Establish the initial inequality.

We know that the cosine function has a fixed range. For any angle $\theta$, $-1\le \cos (\theta )\le 1$. This holds true even for a complex angle like $\frac{1}{x}$ (for $x\ne 0$).

So, we can state:

$$-1\le \cos \left(\frac{1}{x}\right)\le 1$$

Step 2: Build the target function.

Our target function is $x^2\cos \left(\frac{1}{x}\right)$. To get this expression in the middle of our inequality, we must multiply all three parts by $x^2$. Since $x^2\ge 0$ for all $x$, the direction of the inequality symbols will not change.

$$-1\cdot x^2\le x^2\cos \left(\frac{1}{x}\right)\le 1\cdot x^2$$

$$-x^2\le x^2\cos \left(\frac{1}{x}\right)\le x^2$$

Step 3: Identify the bounding functions and find their limits.

We have trapped our target function between $g(x)=-x^2$ and $h(x)=x^2$. Now, we find their limits as $x\to 0$.

• Limit of the lower bound: ${\lim }_{x\to 0}g(x)={\lim }_{x\to 0}(-x^2)=-(0{)}^2=0$

• Limit of the upper bound: ${\lim }_{x\to 0}h(x)={\lim }_{x\to 0}(x^2)=(0{)}^2=0$

Step 4: Apply the Squeeze Theorem.

Since $-x^2\le x^2\cos \left(\frac{1}{x}\right)\le x^2$ and ${\lim }_{x\to 0}(-x^2)={\lim }_{x\to 0}(x^2)=0$, we can conclude by the Squeeze Theorem that:

$$\underset{x\to 0}{\lim }{x}^2\cos \left(\frac{1}{x}\right)=0$$

Step-by-Step Example 2: Exam-Style Application

Problem: Let $f(x)$ be a function such that $2x-1\le f(x)\le x^2-2x+3$ for all $x$. What is ${\lim }_{x\to 2}f(x)$?

Solution:

In this problem, the inequality is given to us directly. We do not need to derive it. Our task is to verify the conditions of the Squeeze Theorem and apply it.

Step 1: Identify the bounding functions.

The problem explicitly states the bounds.

• Lower bound: $g(x)=2x-1$

• Upper bound: $h(x)=x^2-2x+3$

Step 2: Calculate the limit of the lower bound as $x\to 2$.

The function $g(x)$ is a polynomial, so we can find its limit by direct substitution.

$$\underset{x\to 2}{\lim }g(x)=\underset{x\to 2}{\lim }(2x-1)=2(2)-1=4-1=3$$

Step 3: Calculate the limit of the upper bound as $x\to 2$.

The function $h(x)$ is also a polynomial, so we can use direct substitution.

$$\underset{x\to 2}{\lim }h(x)=\underset{x\to 2}{\lim }(x^2-2x+3)=(2{)}^2-2(2)+3=4-4+3=3$$

Step 4: Apply the Squeeze Theorem.

We have shown that ${\lim }_{x\to 2}g(x)=3$ and ${\lim }_{x\to 2}h(x)=3$. Since the limits of the upper and lower bounds are equal, and $f(x)$ is trapped between them, we can conclude by the Squeeze Theorem that the limit of $f(x)$ must also be the same.

$$\underset{x\to 2}{\lim }f(x)=3$$

Using Your Calculator

The Squeeze Theorem is a purely analytical technique; you cannot "use the Squeeze Theorem" on a calculator. However, a graphing calculator is an excellent tool for visualizing why the theorem works and for confirming your answer.

To visualize the result from Example 1 (${\lim }_{x\to 0}{x}^2\cos \left(\frac{1}{x}\right)=0$):

1. Press the Y= button.

2. In $Y_1$, enter the lower bound: $Y_1=-X^2$

3. In $Y_2$, enter the target function: $Y_2=X^2\ast cos(1/X)$

4. In $Y_3$, enter the upper bound: $Y_3=X^2$

5. Press ZOOM and select 6:ZStandard. You will see the general shape.

6. To see the "squeeze" near $x=0$, press ZOOM and select 2:Zoom In. Center the cursor at the origin $(0,0)$ and press ENTER.

7. You will now see a clear visual representation of the function $Y_2$ oscillating but being "squeezed" between the parabolas $Y_1$ and $Y_3$ as they all converge at the point $(0,0)$. This provides graphical confirmation of your analytical result.

AP Exam Quick Hit

Common Question Types

• Building the Inequality: You are asked to find a limit like ${\lim }_{x\to 0}g(x)\cdot \sin (1/x)$. You are expected to know that $-1\le \sin (1/x)\le 1$ and use it to construct the full inequality before applying the theorem.

• Using a Given Inequality: A multiple-choice or free-response question provides the full inequality, such as "If $h(x)\le f(x)\le g(x)$ for all $x$ and ${\lim }_{x\to a}h(x)={\lim }_{x\to a}g(x)=5$, what is ${\lim }_{x\to a}f(x)$?". This is a direct test of your understanding of the theorem's conclusion.

• Justification: In a free-response question, you may be asked to "Show that ${\lim }_{x\to c}f(x)=L$". If the Squeeze Theorem is the intended method, you must provide a full justification by stating the inequality, showing the work for calculating the limits of the two outer functions, and explicitly stating that because those limits are equal, the limit of the center function must be the same by the Squeeze Theorem.

Common Mistakes

• Incorrect Initial Inequality: Starting with an incorrect range for a trigonometric function, such as $0\le \sin (x)\le 1$, which is only true for certain intervals, not for all $x$. The correct universal range is $-1\le \sin (x)\le 1$.

• Algebraic Errors when Multiplying: When multiplying an inequality by a variable expression (e.g., multiplying by $x$), forgetting to consider that if $x$ is negative, the inequality signs must be flipped. It is safer to multiply by an always-positive quantity like $x^2$ or $|x|$.

• Incomplete Justification: On a free-response question, simply writing down the inequality and then stating the final answer without showing the evaluation of the limits for the two bounding functions. You must explicitly show that $\lim g(x)=L$ and $\lim h(x)=L$.

• Assuming the Theorem Applies: If the limits of the outer functions are not equal (e.g., one is 3 and the other is 5), the Squeeze Theorem cannot be used. A common mistake is to try to "average" the limits or guess. The correct conclusion is that the theorem is not applicable.