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

The Core Idea: Determining Limits Using the Squeeze Theorem

The Squeeze Theorem provides a powerful analytical method for determining the limit of a function that may be difficult to evaluate directly, especially those involving trigonometric oscillations. The fundamental concept is to find the limit of a function by "trapping" or "squeezing" it between two other functions whose limits are more easily determined. If these two "bounding" functions approach the same limit at a particular point, then the function trapped between them must also approach that same limit.

This theorem is particularly useful for finding limits of functions that are products of a term approaching zero and a bounded, oscillating function. It allows us to make a definitive conclusion about the limit of the trapped function by analyzing the behavior of the simpler functions that form its upper and lower bounds. The theorem serves as the formal justification for certain fundamental trigonometric limits that are foundational in calculus.

The Squeeze Theorem

The Squeeze Theorem is a formal statement about the limit of a function that is bounded by two other functions.

The theorem states:

If functions $f$, $g$, and $h$ satisfy the inequality $g(x)\le f(x)\le h(x)$ for all $x$ in an 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

To successfully apply the Squeeze Theorem, three distinct conditions must be met and justified. Failure to establish any one of these conditions invalidates the conclusion.

1. The Inequality Condition: There must be an established inequality, $g(x)\le f(x)\le h(x)$, that holds true for all $x$ in an open interval around the point $c$. The behavior of the functions at$x=c$ is not relevant to the condition, only the behavior near$c$. This inequality is often constructed by using the known bounds of a part of the function, such as $-1\le \sin (x)\le 1$ or $-1\le \cos (x)\le 1$.

2. The Equal Limits Condition: The limits of the two outer, bounding functions, $g(x)$ and $h(x)$, must both exist as $x$ approaches $c$, and they must be equal to the same finite value, $L$. If ${\lim }_{x\to c}g(x)\ne {\lim }_{x\to c}h(x)$, the Squeeze Theorem cannot be used to draw a conclusion about the limit of $f(x)$.

3. The Conclusion: If and only if the first two conditions are met, you can conclude that the limit of the inner function, $f(x)$, is also equal to $L$.

A primary application of this theorem, as specified in the AP curriculum, is to provide the rigorous proof for important limits like ${\lim }_{x\to 0}\frac{\sin (x)}{x}=1$. While you may not be asked to reproduce the geometric proof, you are expected to understand that the Squeeze Theorem is the tool that justifies this result.

Core Concepts & Rules

• The Squeeze Theorem is a method for finding the limit of a function, $f(x)$, by trapping it between two other functions, $g(x)$ and $h(x)$.

• To use the theorem, you must first establish an inequality of the form $g(x)\le f(x)\le h(x)$ that is valid on an open interval containing the point of interest, $c$.

• You must then show that the limits of the lower and upper bounding functions are equal: ${\lim }_{x\to c}g(x)={\lim }_{x\to c}h(x)=L$.

• If both the inequality and the equal limits conditions are satisfied, you can conclude that the limit of the trapped function is also $L$: ${\lim }_{x\to c}f(x)=L$.

• The Squeeze Theorem is the formal justification for the fundamental trigonometric limit ${\lim }_{x\to 0}\frac{\sin (x)}{x}=1$.

Step-by-Step Example 1: A Function with a Trigonometric Component

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

Step 1: Analyze the function and identify the bounded part.

The function is $f(x)=x^2\cos \left(\frac{1}{x}\right)$. Direct substitution of $x=0$ results in an undefined expression inside the cosine function. We can see that as $x\to 0$, the term $\cos \left(\frac{1}{x}\right)$ oscillates infinitely between -1 and 1, while the $x^2$ term approaches 0. This structure is a prime candidate for the Squeeze Theorem. The bounded part is the cosine function.

Step 2: Establish the initial inequality.

We know that the cosine function is always bounded between -1 and 1, regardless of its input.

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

Step 3: Build the full function within the inequality.

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

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

Step 4: Identify the bounding functions and find their limits.

Our inequality is in the form $g(x)\le f(x)\le h(x)$.

• The lower bound is $g(x)=-x^2$.

• The upper bound is $h(x)=x^2$.

Now, we find the limits of $g(x)$ and $h(x)$ as $x\to 0$.

$$\underset{x\to 0}{\lim }g(x)=\underset{x\to 0}{\lim }(-x^2)=-(0{)}^2=0$$

$$\underset{x\to 0}{\lim }h(x)=\underset{x\to 0}{\lim }(x^2)=(0{)}^2=0$$

Step 5: Apply the Squeeze Theorem to state the conclusion.

We have successfully shown that $-x^2\le x^2\cos \left(\frac{1}{x}\right)\le x^2$ and that ${\lim }_{x\to 0}(-x^2)={\lim }_{x\to 0}(x^2)=0$.

Therefore, by the Squeeze Theorem, we can conclude:

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

Step-by-Step Example 2: Justifying a Limit from a Given Inequality

Problem: It is known that for functions $f$, $g$, and $h$, the inequality $h(x)\le f(x)\le g(x)$ holds for all $x$. The graphs of $g$ and $h$ are shown below. Determine ${\lim }_{x\to 2}f(x)$ and justify your reasoning.

(Imagine a graph where both $g(x)$ and $h(x)$ are continuous functions that meet at the point $(2,3)$. For example, $g(x)$ could be a parabola opening up and $h(x)$ could be a parabola opening down, both with a vertex at $(2,3)$.)

Step 1: State the given inequality.

We are given that $h(x)\le f(x)\le g(x)$ for all $x$.

Step 2: Determine the limits of the bounding functions from the graph.

We need to find the limits of $h(x)$ and $g(x)$ as $x$ approaches 2. By inspecting the provided graph:

• As $x$ approaches 2 from the left and the right, the $y$-values of the graph of $h(x)$ approach 3. Therefore, ${\lim }_{x\to 2}h(x)=3$.

• As $x$ approaches 2 from the left and the right, the $y$-values of the graph of $g(x)$ approach 3. Therefore, ${\lim }_{x\to 2}g(x)=3$.

Step 3: Check the conditions for the Squeeze Theorem.

We have satisfied both necessary conditions:

1. The inequality $h(x)\le f(x)\le g(x)$ is given.

2. The limits of the bounding functions are equal: ${\lim }_{x\to 2}h(x)={\lim }_{x\to 2}g(x)=3$.

Step 4: State the conclusion based on the Squeeze Theorem.

Because both conditions are met, we can apply the Squeeze Theorem.

Therefore, ${\lim }_{x\to 2}f(x)=3$.

Using Your Calculator

The Squeeze Theorem is a purely analytical tool; its application and justification cannot be done on a calculator. You must write out the logical steps (the inequality and the limits of the bounding functions) to receive credit on the AP Exam.

However, a graphing calculator is an excellent tool for visualizing the concept and verifying your answer. To check the result from Example 1, ${\lim }_{x\to 0}{x}^2\cos \left(\frac{1}{x}\right)=0$:

1. Press Y=.

2. Enter the three functions:

   • $Y_1=-X^2$ (the lower bound, $g(x)$)

   • $Y_2=X^{2}cos(1/X)$ (the function in question, $f(x)$)

   • $Y_3=X^2$ (the upper bound, $h(x)$)

3. Press ZOOM and select 4:ZDecimal or 6:ZStandard. You may need to adjust the window manually to zoom in closer to the origin (e.g., $Xmin=-1$, $Xmax=1$, $Ymin=-1$, $Ymax=1$).

4. Observe the graph. You will see the graph of $Y_2$ oscillating between the graphs of $Y_1$ and $Y_3$. As the graphs converge toward $x=0$, you can see $Y_2$ being "squeezed" between the other two functions, all heading toward a $y$-value of 0.

5. You can also use the table feature (2ND + GRAPH). Set the table to start at 0 with a very small step (e.g., 0.001) to see numerically that all three functions approach 0 as $x$ gets close to 0.

AP Exam Quick Hit

Common Question Types

• Direct Analytical Application: You will be given a function, typically a product of a polynomial and a sine or cosine function, such as $f(x)=(x-1{)}^2\sin \left(\frac{1}{x-1}\right)$, and asked to find ${\lim }_{x\to 1}f(x)$. You must construct the inequality and apply the theorem for full credit.

• Justification with Given Information: You will be told that $g(x)\le f(x)\le h(x)$ and be provided with information about $g$ and $h$ in the form of equations, a graph, or a table of values. You will then be asked to find the limit of $f(x)$ at a specific point, requiring you to assemble the pieces into a justification using the Squeeze Theorem.

Common Mistakes

• Incomplete Justification: Simply stating the final answer for the limit without showing the full argument. A complete justification requires explicitly stating the inequality, showing that the limits of the outer functions are equal, and then invoking the Squeeze Theorem.

• Incorrect Bounds for Trig Functions: Using incorrect bounds for sine or cosine. For example, using $0\le \sin (x)\le 1$ when the correct bound is $-1\le \sin (x)\le 1$. This will lead to an incorrect initial inequality.

• Algebraic Errors when Multiplying: Forgetting to multiply all three parts of the inequality by the same term, or incorrectly handling the inequality signs if multiplying by a term that could be negative.

• Applying the Theorem when Limits are Unequal: Attempting to use the Squeeze Theorem when ${\lim }_{x\to c}g(x)\ne {\lim }_{x\to c}h(x)$. In this case, the theorem is not applicable and no conclusion can be drawn about ${\lim }_{x\to c}f(x)$.