The Squeeze Theorem
The **Squeeze Theorem**, also known as the **Sandwich Theorem**, is a powerful tool in calculus for finding the limit of a function that is "squeezed" between two other functions. The idea is simple: if a function \(h(x)\) is always between two other functions, \(f(x)\) and \(g(x)\), and both \(f(x)\) and \(g(x)\) approach the same limit \(L\) at a certain point, then \(h(x)\) must also approach that same limit \(L\).
The theorem is formally stated as follows:
Let \(f(x)\), \(g(x)\), and \(h(x)\) be three functions such that for all \(x\) in an open interval containing \(c\) (except possibly at \(c\) itself):
\[ f(x) \le h(x) \le g(x) \]If
\[ \lim_{x \to c} f(x) = L \quad \text{and} \quad \lim_{x \to c} g(x) = L \]Then
\[ \lim_{x \to c} h(x) = L \]This is especially useful for functions that involve sine or cosine, as their values are always bounded between -1 and 1. The key to applying the theorem is to create the bounding functions, \(f(x)\) and \(g(x)\), from the original inequality.
---Practice Examples
Example 1: Basic Trigonometric Limit
Problem: Find \( \lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) \).
Solution: We know that for any value of \(\theta\), \( -1 \le \sin(\theta) \le 1 \). Therefore, we can say that \( -1 \le \sin\left(\frac{1}{x}\right) \le 1 \).
Now, we need to build the function \( x^2 \sin\left(\frac{1}{x}\right) \) from this inequality. We multiply all parts of the inequality by \( x^2 \). Since \( x^2 \) is always greater than or equal to 0, the direction of the inequality signs does not change.
\[ -x^2 \le x^2 \sin\left(\frac{1}{x}\right) \le x^2 \]Next, we find the limits of the two "squeezing" functions as \( x \) approaches 0.
\[ \lim_{x \to 0} (-x^2) = 0 \] \[ \lim_{x \to 0} (x^2) = 0 \]Since both the lower and upper bounds approach 0, by the Squeeze Theorem, the limit of the function in the middle must also be 0.
\[ \lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) = 0 \] ---Check Yourself Problems
Apply the Squeeze Theorem to solve each problem below. For each problem, find the bounding functions and their limits to determine the limit of the given function.
Problem 1: Find the limit using the Squeeze Theorem: \( \lim_{x \to 0} x^4 \cos\left(\frac{1}{x}\right) \).
We know that \( -1 \le \cos\left(\frac{1}{x}\right) \le 1 \). Multiplying by \(x^4\), we get:
\[ -x^4 \le x^4 \cos\left(\frac{1}{x}\right) \le x^4 \]Since \( \lim_{x \to 0} (-x^4) = 0 \) and \( \lim_{x \to 0} x^4 = 0 \), by the Squeeze Theorem, the limit is 0.
\[ \lim_{x \to 0} x^4 \cos\left(\frac{1}{x}\right) = 0 \]Problem 2: Find the limit using the Squeeze Theorem: \( \lim_{x \to \infty} \frac{\cos(x^2)}{x} \).
We know that \( -1 \le \cos(x^2) \le 1 \). For \(x > 0\), we can divide by \(x\):
\[ -\frac{1}{x} \le \frac{\cos(x^2)}{x} \le \frac{1}{x} \]Since \( \lim_{x \to \infty} \left(-\frac{1}{x}\right) = 0 \) and \( \lim_{x \to \infty} \left(\frac{1}{x}\right) = 0 \), by the Squeeze Theorem, the limit is 0.
\[ \lim_{x \to \infty} \frac{\cos(x^2)}{x} = 0 \]Problem 3: Find the limit using the Squeeze Theorem: \( \lim_{x \to 0} \left(x \sin\left(\frac{1}{x}\right)\right) \).
We know that \( -1 \le \sin\left(\frac{1}{x}\right) \le 1 \). Multiplying by \(|x|\), we get:
\[ -|x| \le x \sin\left(\frac{1}{x}\right) \le |x| \]Since \( \lim_{x \to 0} (-|x|) = 0 \) and \( \lim_{x \to 0} |x| = 0 \), by the Squeeze Theorem, the limit is 0.
\[ \lim_{x \to 0} \left(x \sin\left(\frac{1}{x}\right)\right) = 0 \]Problem 4: Given that \( 3x-1 \le f(x) \le x^2-3x+8 \) for all \(x\), find \( \lim_{x \to 3} f(x) \).
We find the limits of the bounding functions as \(x \to 3\):
- Lower bound: \( \lim_{x \to 3} (3x-1) = 3(3)-1 = 8 \)
- Upper bound: \( \lim_{x \to 3} (x^2-3x+8) = 3^2-3(3)+8 = 9-9+8 = 8 \)
Since both limits are 8, by the Squeeze Theorem, the limit of \(f(x)\) is also 8.
\[ \lim_{x \to 3} f(x) = 8 \]Problem 5: Find the limit using the Squeeze Theorem: \( \lim_{x \to 0} x \cos\left(\frac{5}{x}\right) \).
We know that \( -1 \le \cos\left(\frac{5}{x}\right) \le 1 \). Multiplying by \(|x|\), we get:
\[ -|x| \le x \cos\left(\frac{5}{x}\right) \le |x| \]Since \( \lim_{x \to 0} (-|x|) = 0 \) and \( \lim_{x \to 0} |x| = 0 \), by the Squeeze Theorem, the limit is 0.
\[ \lim_{x \to 0} x \cos\left(\frac{5}{x}\right) = 0 \]