The Squeeze Theorem, also known as the Sandwich Theorem, plays a pivotal role in calculus for determining the limits of functions that are difficult to evaluate directly. It relies on the concept of "squeezing" a function between two others to find its limit.

📚 Introduction to the Squeeze Theorem

The theorem is founded on a simple yet powerful premise: if a function $f(x)$ is always caught between two other functions $g(x)$ and $h(x)$ near a certain point, and the limits of $g(x)$ and $h(x)$ at that point are equal, then the limit of $f(x)$ at that point must exist and be equal to the same value. This principle is crucial for dealing with functions that oscillate or approach indeterminate forms.

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💡 Prerequisites for Applying the Squeeze Theorem

• Functions $g(x)$ and $h(x)$ must bound $f(x)$ near the point of interest.
• Limits of $g(x)$ and $h(x)$ at the point must exist and be equal.

💡 Significance in Calculus

• Dealing with Indeterminate Forms: Offers a method to evaluate limits of functions that are otherwise challenging to approach directly.
• Understanding Oscillations: Enables the calculation of limits for functions that oscillate within a certain range.

Worked Examples

🔍 Example 1: Evaluating a Basic Limit

Consider the functions $g(x) = x^2), (f(x) = x^2 \sin(\frac{1}{x})$ for $x \neq 0),$, and $h(x) = -x^2$ and evaluate $\lim_{x \to 0} f(x)$.

• Step 1: Establish the inequalities $h(x) \leq f(x) \leq g(x)$ for all $x$ near 0.
• Step 2: Calculate the limits of $g(x)$ and $h(x)$ as $x$ approaches 0.

$\begin{aligned} \lim_{x \to 0} g(x) &= \lim_{x \to 0} x^2 = 0, \\ \lim_{x \to 0} h(x) &= \lim_{x \to 0} -x^2 = 0. \end{aligned}$<p></p><ul><li><strong>Step 3:</strong> Apply the Squeeze Theorem to conclude that$\lim_{x \to 0} f(x) = 0$.</li></ul><h3>🔍 <strong>Example 2: Oscillating Function</strong></h3><p>Evaluate$\lim_{x \to 0} x \sin(\frac{1}{x})$.</p><ul><li><strong>Step 1:</strong> Identify the bounding functions. Here,$g(x) = x$and$h(x) = -x$.</li><li><strong>Step 2:</strong> Note that for all$x \neq 0, -x \leq x \sin(\frac{1}{x}) \leq x$.</li><li><strong>Step 3:</strong> Calculate the limits of the bounding functions as$x$approaches 0.</li></ul><p></p>$\begin{aligned} \lim_{x \to 0} (-x) &= 0, \\ \lim_{x \to 0} x &= 0. \end{aligned}$<p></p><ul><li><strong>Step 4:</strong> By the Squeeze Theorem,$\lim_{x \to 0} x \sin(\frac{1}{x}) = 0$.</li></ul><p>These examples illustrate the theorem's utility in resolving limits that are not straightforward to evaluate directly. By setting up appropriate bounding functions and applying the Squeeze Theorem, students can tackle a wide range of problems involving limits.</p><h2 id="practice-questions">✏️ <strong>Practice Questions</strong></h2><h3>📝 <strong>Question 1</strong></h3><p>Evaluate the limit$\lim_{x \to 0} x^2 \cos(\dfrac{1}{x})$.</p><p></p><h3>📝 <strong>Question 2</strong></h3><p>Determine the limit$\lim_{x \to 0} x^4 \sin(\dfrac{1}{x})$.</p><p></p><h2 id="solutions-to-practice-questions">✅ <strong>Solutions to Practice Questions</strong></h2><h3>🧩 <strong>Solution to Question 1</strong></h3><p><strong>Step 1:</strong> Identify the function to be evaluated:$f(x) = x^2 \cos(\frac{1}{x})$for$x \neq 0$.</p><p><strong>Step 2:</strong> Choose appropriate bounding functions. Since$-1 \leq \cos(\frac{1}{x}) \leq 1$for all$x$, we can use$g(x) = -x^2$and$h(x) = x^2$as the lower and upper bounds, respectively.</p><p><strong>Step 3:</strong> Evaluate the limits of the bounding functions as$x$approaches 0.</p><p></p>$\begin{aligned} \lim_{x \to 0} g(x) &= \lim_{x \to 0} -x^2 = 0, \\ \lim_{x \to 0} h(x) &= \lim_{x \to 0} x^2 = 0. \end{aligned}$<p></p><p><strong>Step 4:</strong> Apply the Squeeze Theorem. Since$g(x) \leq f(x) \leq h(x)$and both$g(x)$and$h(x)$approach 0 as$x$approaches 0, by the Squeeze Theorem,$f(x)$also approaches 0.</p><p></p>$\lim_{x \to 0} x^2 \cos(\frac{1}{x}) = 0.$<p></p><h3>🧩 <strong>Solution to Question 2</strong></h3><p><strong>Step 1:</strong> Consider the function$f(x) = x^4 \sin(\frac{1}{x}))$for$x \neq 0$.</p><p><strong>Step 2:</strong> For bounding functions, use$g(x) = -x^4$and$h(x) = x^4$, leveraging the fact that$-1 \leq \sin(\frac{1}{x}) \leq 1$.</p><p><strong>Step 3:</strong> Calculate the limits of$g(x)$and$h(x)$as$x$approaches 0.</p><p></p>$\begin{aligned} \lim_{x \to 0} g(x) &= \lim_{x \to 0} -x^4 = 0, \\ \lim_{x \to 0} h(x) &= \lim_{x \to 0} x^4 = 0. \end{aligned}$<p></p><p><strong>Step 4:</strong> By the Squeeze Theorem, since$g(x) \leq f(x) \leq h(x)$and the limits of$g(x)$and$h(x)$as$x$approaches 0 are both 0, the limit of$f(x)$as$x$approaches 0 must also be 0.</p><p></p>$\lim_{x \to 0} x^4 \sin(\frac{1}{x}) = 0.$