1. The Idea of the Squeeze Theorem

The **Squeeze Theorem** is a powerful tool used to find the limit of a function that is difficult to evaluate directly. The idea is to "squeeze" the function between two other functions that have a known, and equal, limit at a specific point. If the "squeezing" functions both approach the same value, the function in the middle must also approach that same value.

This theorem is particularly useful for limits involving trigonometric functions, such as sine and cosine, which oscillate between -1 and 1.

2. Formal Definition

Let \( f(x) \), \( g(x) \), and \( h(x) \) be functions such that \( f(x) \le g(x) \le h(x) \) for all \( x \) in some open interval containing \( c \), except possibly at \( c \) itself. If \[ \lim_{x \to c} f(x) = L \] and \[ \lim_{x \to c} h(x) = L \] then the limit of the middle function also exists and is equal to \( L \): \[ \lim_{x \to c} g(x) = L \]

3. Applying the Theorem

The most common application of the Squeeze Theorem is with functions like \( g(x) = x^2 \sin(\frac{1}{x}) \). We know that the value of \( \sin(\frac{1}{x}) \) is always between -1 and 1, inclusive. We can write this as:

\[ -1 \le \sin\left(\frac{1}{x}\right) \le 1 \]

Now, we can multiply all parts of the inequality by \( x^2 \). Since \( x^2 \) is always non-negative, the inequality signs do not flip.

\[ -x^2 \le x^2 \sin\left(\frac{1}{x}\right) \le x^2 \]

Next, we evaluate the limits of the two "squeezing" functions as \( x \to 0 \).

\[ \lim_{x \to 0} (-x^2) = 0 \] \[ \lim_{x \to 0} (x^2) = 0 \]

Since both the lower and upper functions approach 0, the Squeeze Theorem guarantees that the function in the middle must also approach 0.

\[ \lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) = 0 \]

4. Check Yourself

Question 1: Use the Squeeze Theorem to evaluate \( \lim_{x \to 0} x^4 \cos\left(\frac{3}{x}\right) \).

We know that \( -1 \le \cos\left(\frac{3}{x}\right) \le 1 \).

Multiply all parts by \( x^4 \) (which is always non-negative):

\[ -x^4 \le x^4 \cos\left(\frac{3}{x}\right) \le x^4 \]

Now, take the limit of the outer functions as \( x \to 0 \):

\[ \lim_{x \to 0} (-x^4) = 0 \] \[ \lim_{x \to 0} (x^4) = 0 \]

Since both limits are equal to 0, by the Squeeze Theorem, the limit of the middle function must also be 0.

\[ \lim_{x \to 0} x^4 \cos\left(\frac{3}{x}\right) = 0 \]