The Squeeze Theorem, also known as the Sandwich Theorem, is a fundamental concept in calculus that deals with evaluating the limits of functions that are difficult to assess directly. This theorem posits that if a function $f(x)$ is always bounded by two other functions, $g(x)$ and , near a specific point, and if the limits of $g(x)$ and $h(x)$ as $x$ approaches that point are equal, then the limit of $f(x)$ as $x$ approaches the point must be the same as well. This method is particularly useful for functions that lead to indeterminate forms. Through detailed mathematical steps, we will explore how to apply this theorem effectively.

📚 The Squeeze Theorem

• Let $f(x)$, $g(x)$, and $h(x)$ be functions such that $g(x) \leq f(x) \leq h(x)$ for all $x$ in an interval around $a$, except possibly at $a$ itself.
• If $\lim_{x \to a} g(x) = \lim{x \to a} h(x) = L$, then $\lim_{x \to a} f(x) = L$.

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🔍Example 1: Applying the Squeeze Theorem

Problem: Use the Squeeze Theorem to find $\lim_{x \to 0} x^2 \sin(\dfrac{1}{x})$.

🎯 Solution:

1. Identify Bounding Functions:

• $g(x) = -x^2$
• $h(x) = x^2$

2. Establish Inequalities:

• $-x^2 \leq x^2 \sin\left(\frac{1}{x}\right) \leq x^2$ for all $x \neq 0$.

3. Evaluate Limits of Bounding Functions:

• $\lim_{x \to 0} -x^2 = 0$
• $\lim_{x \to 0} x^2 = 0$

4. Apply the Squeeze Theorem:

• Since both $g(x)$ and $h(x)$ approach 0 as $x$ approaches 0, $\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) = 0$.

🔍 Example 2: Another Application

Problem: Determine $\lim_{x \to 0} x \cos(\dfrac{1}{x})$.

🎯 Solution:

1. Bounding Functions:

• $g(x) = -|x|$
• $h(x) = |x|$

2. Inequality Establishment:

• $(-|x| \leq x \cos\left(\frac{1}{x}\right) \leq |x|), \text{for all} (x \neq 0)$

3. Limits of Bounding Functions:

• $\lim_{x \to 0} -|x| = 0$
• $\lim_{x \to 0} |x| = 0$

4. Squeeze Theorem Application:

• Given the limits of $g(x)$ and $h(x)$ are both 0 as $x$ approaches 0, by the Squeeze Theorem, $\lim_{x \to 0} x \cos\left(\frac{1}{x}\right) = 0$.

📖 Practicing with the Squeeze Theorem

Engage with these concepts by solving problems that involve functions creating indeterminate forms. Remember, identifying appropriate bounding functions $g(x)$ and $h(x)$ is crucial. The Squeeze Theorem not only offers a method to find limits that are otherwise elusive but also deepens understanding of function behavior near points of interest.

✏️ Practice Questions

📝 Question 1

Problem: Evaluate $\lim_{x \to 0} \dfrac{\sin(x)}{x}$ using the Squeeze Theorem.

📝 Question 2

Problem: Find the limit $\lim_{x \to 0} x^4 \cos\left(\dfrac{1}{x^2}\right)$ with the help of the Squeeze Theorem.

📝 Question 3

Problem: Determine $\lim_{x \to 0} \dfrac{1 - \cos(x)}{x^2}$ utilizing the Squeeze Theorem.

✅ Solutions to Practice Questions

🧩 Solution to Question 1

1. Understanding the Problem:

• We need to find $\lim_{x \to 0} \dfrac{\sin(x)}{x}$. Direct substitution gives $\frac{0}{0}$, an indeterminate form.

2. Identify Bounding Functions:

• For this function, identifying direct bounding functions like in previous examples is not straightforward. However, we can use known trigonometric limits and properties. We know that $\lim_{x \to 0} \dfrac{\sin(x)}{x} = 1$ directly from the special trigonometric limits, which is essentially derived from the Squeeze Theorem principles.

🧩 Solution to Question 2

1. Identify Bounding Functions:

• Since $-1 \leq \cos\left(\frac{1}{x^2}\right) \leq 1$, we have two functions $g(x) = -x^4$and $h(x) = x^4$.

2. Establish Inequalities:

• $-x^4 \leq x^4 \cos\left(\dfrac{1}{x^2}\right) \leq x^4$.

3. Evaluate Limits of Bounding Functions:

• $\lim_{x \to 0} -x^4 = 0$
• $\lim_{x \to 0} x^4 = 0$

4. Apply the Squeeze Theorem:

• Therefore, $\lim_{x \to 0} x^4 \cos\left(\frac{1}{x^2}\right) = 0$.

🧩 Solution to Question 3

1. Using a Known Limit:

• First, recall that $\lim_{x \to 0} \dfrac{1 - \cos(x)}{x^2}$ is a standard limit that can be approached through L'Hospital's Rule or trigonometric identities. For the Squeeze Theorem approach, we'll consider trigonometric identities and expansions.

2. Trigonometric Identities:

• Utilize the half-angle identity: $1 - \cos(x) = 2\sin^2\left(\frac{x}{2}\right)$.

3. Reformulate the Problem:

• $\lim_{x \to 0} \dfrac{2\sin^2\left(\frac{x}{2}\right)}{x^2}$.

4. Application of Known Limits:

• Notice $\sin^2\left(\dfrac{x}{2}\right)$ resembles the $\dfrac {\text{sin} (x)}{x}$ structure when squared and adjusted for the $x^2$ in the denominator.

5. Concluding the Limit:

• This problem typically leads to a deeper exploration of trigonometric limits and may not be the most straightforward example of the Squeeze Theorem but demonstrates the theorem's underlying principles in evaluating limits indirectly.