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.

Image courtesy of Calcworkshop

💡 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.

$\begin{aligned} \lim_{x \to 0} g(x) &= \lim_{x \to 0} x^2 = 0, \\ \lim_{x \to 0} h(x) &= \lim_{x \to 0} -x^2 = 0. \end{aligned}$

• Step 3: Apply the Squeeze Theorem to conclude that $\lim_{x \to 0} f(x) = 0$.

🔍 Example 2: Oscillating Function

Evaluate $\lim_{x \to 0} x \sin(\frac{1}{x})$.

• Step 1: Identify the bounding functions. Here, $g(x) = x$ and $h(x) = -x$.

• Step 2: Note that for all $x \neq 0, -x \leq x \sin(\frac{1}{x}) \leq x$.

• Step 3: Calculate the limits of the bounding functions as $x$ approaches 0.

$\begin{aligned} \lim_{x \to 0} (-x) &= 0, \\ \lim_{x \to 0} x &= 0. \end{aligned}$

• Step 4: By the Squeeze Theorem, $\lim_{x \to 0} x \sin(\frac{1}{x}) = 0$.

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.

✏️ Practice Questions

📝 Question 1

Evaluate the limit $\lim_{x \to 0} x^2 \cos(\dfrac{1}{x})$.

📝 Question 2

Determine the limit $\lim_{x \to 0} x^4 \sin(\dfrac{1}{x})$.

✅ Solutions to Practice Questions

🧩 Solution to Question 1

Step 1: Identify the function to be evaluated: $f(x) = x^2 \cos(\frac{1}{x})$ for $x \neq 0$.

Step 2: 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.

Step 3: Evaluate the limits of the bounding functions as $x$ approaches 0.

$\begin{aligned} \lim_{x \to 0} g(x) &= \lim_{x \to 0} -x^2 = 0, \\ \lim_{x \to 0} h(x) &= \lim_{x \to 0} x^2 = 0. \end{aligned}$

Step 4: 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.

$\lim_{x \to 0} x^2 \cos(\frac{1}{x}) = 0.$

🧩 Solution to Question 2

Step 1: Consider the function $f(x) = x^4 \sin(\frac{1}{x}))$ for $x \neq 0$.

Step 2: For bounding functions, use $g(x) = -x^4$ and $h(x) = x^4$, leveraging the fact that $-1 \leq \sin(\frac{1}{x}) \leq 1$.

Step 3: Calculate the limits of $g(x)$ and $h(x)$ as $x$ approaches 0.

$\begin{aligned} \lim_{x \to 0} g(x) &= \lim_{x \to 0} -x^4 = 0, \\ \lim_{x \to 0} h(x) &= \lim_{x \to 0} x^4 = 0. \end{aligned}$

Step 4: 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.

$\lim_{x \to 0} x^4 \sin(\frac{1}{x}) = 0.$

Other:

Final stored value at location 10: 8

# Simple simulation of the Fetch-Execute Cycle

# Example program stored in memory (a list of instructions)
memory = [
    "LOAD 5",     # Load the number 5 into the accumulator
    "ADD 3",      # Add 3 to the accumulator
    "STORE 10",   # Store the result at memory location 10
]

# Simple registers
program_counter = 0    # Holds the address of the next instruction
accumulator = 0        # Holds the result of calculations
memory_store = [0] * 20  # Create a simple memory store with 20 locations

while program_counter < len(memory):
    # FETCH
    instruction = memory[program_counter]

    # DECODE
    parts = instruction.split()
    operation = parts[0]
    value = int(parts[1])

    # EXECUTE
    if operation == "LOAD":
        accumulator = value
    elif operation == "ADD":
        accumulator += value
    elif operation == "STORE":
        memory_store[value] = accumulator

    # Move to the next instruction
    program_counter += 1

# Output final result stored in memory
print("Final stored value at location 10:", memory_store[10])