#Study Notes: Squeeze Theorem and Trigonometric Limits

#Table of Contents

1. Squeeze Theorem
   • Definition
   • Formal Statement
   • Worked Example
2. Trigonometric Limits
   • Important Trigonometric Limit Theorems
   • Worked Examples
3. Practice Questions
4. Glossary
5. Summary and Key Takeaways

#Squeeze Theorem

#Definition

The squeeze theorem is a method used to determine the limit of a function that is bounded above and below by two other functions. If these bounding functions converge to the same limit at a specific point, the bounded function will also converge to that same limit.

#Formal Statement

Let $f$, $g$, and $h$ be functions defined on an open interval containing $a$ such that:

• $g(x) \le f(x) \le h(x)$ for all $x$ in the interval, and
• $\lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L$

Then: $$\lim_{x \to a} f(x) = L$$

#Worked Example

Let $f(x) = x^2 - 6x + 13$ and $g(x) = 6x - x^2 - 5$. It is known that $g(x) \le f(x)$ for $0 < x < 6$. Let $h(x)$ be another function such that $0 < x < 6$. Find $\lim_{x \to 3} h(x)$.

Solution: First, find the limits for $f(x)$ and $g(x)$ at $x = 3$ using substitution: $$\lim_{x \to 3} f(x) = (3)^2 - 6(3) + 13 = 4$$ $$\lim_{x \to 3} g(x) = 6(3) - (3)^2 - 5 = 4$$

Since both limits are equal and $g(x) \le h(x) \le f(x)$ holds for $0 < x < 6$, by the squeeze theorem: $$\lim_{x \to 3} h(x) = 4$$

#Trigonometric Limits

#Important Trigonometric Limit Theorems

You should know and be able to use the following trigonometric limit theorems: $$\lim_{x \to 0} \frac{\sin x}{x} = 1$$ $$\lim_{x \to 0} \frac{\cos x - 1}{x} = 0$$

These can be combined with properties of limits and algebraic manipulations to find other limits. Remember the trigonometric identity: $$\sin^2 x + \cos^2 x = 1$$ which can be rearranged to give: $$\cos^2 x = 1 - \sin^2 x$$ or $$\sin^2 x = 1 - \cos^2 x$$

#Worked Examples

#Example 1

Find the limit: $$\lim_{x \to 0} \frac{1 - \cos 3x}{x}$$

Solution: Substitution would give $\frac{0}{0}$, so start with algebraic manipulation: $$\frac{1 - \cos 3x}{x} = \frac{1 - \cos 3x}{x} \cdot \frac{3}{3} = \frac{3(1 - \cos 3x)}{3x}$$

Using the limit theorem: $$\lim_{x \to 0} \frac{\cos 3x - 1}{3x} = 0$$

Thus: $$\lim_{x \to 0} \frac{3(1 - \cos 3x)}{3x} = -3 \cdot 0 = 0$$

#Example 2

Find the limit: $$\lim_{x \to 0} \frac{\sin 7x}{\sin 4x}$$

Solution: Substitution would give $\frac{0}{0}$, so start with algebraic manipulation: $$\frac{\sin 7x}{\sin 4x} = \frac{\sin 7x}{\sin 4x} \cdot \frac{\frac{1}{x}}{\frac{1}{x}} \cdot \frac{\frac{7}{7}}{\frac{4}{4}} = \frac{\left( \frac{7 \sin 7x}{7x} \right)}{\left( \frac{4 \sin 4x}{4x} \right)}$$

Using the limit theorem: $$\lim_{x \to 0} \frac{\sin x}{x} = 1$$

Thus: $$\lim_{x \to 0} \left[ \frac{\left( \frac{7 \sin 7x}{7x} \right)}{\left( \frac{4 \sin 4x}{4x} \right)} \right] = \frac{7}{4} \cdot \frac{\lim_{x \to 0} \frac{\sin 7x}{7x}}{\lim_{x \to 0} \frac{\sin 4x}{4x}} = \frac{7}{4} \cdot \frac{1}{1} = \frac{7}{4}$$

#Example 3

Find the limit: $$\lim_{x \to 0} \frac{1 - \cos x}{x^2}$$

Solution: Substitution would give $\frac{0}{0}$, so start with algebraic manipulation. Using the identity $\sin^2 x = 1 - \cos^2 x$: $$\frac{1 - \cos x}{x^2} = \frac{1 - \cos x}{x^2} \cdot \frac{1 + \cos x}{1 + \cos x} = \frac{1 - \cos^2 x}{x^2 (1 + \cos x)} = \frac{\sin^2 x}{x^2 (1 + \cos x)} = \frac{\left( \frac{\sin x}{x} \right)^2}{1 + \cos x}$$

Using the limit theorem: $$\lim_{x \to 0} \frac{\sin x}{x} = 1$$

Thus: $$\lim_{x \to 0} \left[ \frac{\left( \frac{\sin x}{x} \right)^2}{1 + \cos x} \right] = \frac{\lim_{x \to 0} \left( \frac{\sin x}{x} \right)^2}{\lim_{x \to 0} (1 + \cos x)} = \frac{1^2}{1 + 1} = \frac{1}{2}$$

#Practice Questions

Practice Question

1. Use the squeeze theorem to find $\lim_{x \to 0} x^2 \sin \left( \frac{1}{x} \right)$.

Practice Question

2. Prove the limit $\lim_{x \to 0} \frac{\sin x}{x} = 1$ using the squeeze theorem.

Practice Question

3. Find $\lim_{x \to 0} \frac{\sin 5x}{x}$.

Practice Question

4. Evaluate $\lim_{x \to 0} \frac{x - \sin x}{x^3}$.

#Glossary

• Squeeze Theorem: A method for finding the limit of a function by bounding it between two other functions that have the same limit.
• Trigonometric Limit Theorems: Theorems involving limits of trigonometric functions, such as $\lim_{x \to 0} \frac{\sin x}{x} = 1$.
• Algebraic Manipulation: The process of rearranging and simplifying expressions to make limits easier to evaluate.

#Summary and Key Takeaways

#Summary

• The squeeze theorem is a powerful tool for determining the limits of functions bounded between two other functions.
• Key trigonometric limit theorems are essential for solving limits involving trigonometric functions.
• Algebraic manipulation and trigonometric identities can greatly simplify the process of finding limits.

#Key Takeaways

• Use the squeeze theorem when a function is bounded by two other functions with known limits.
• Remember the important trigonometric limit theorems: $\lim_{x \to 0} \frac{\sin x}{x} = 1$ and $\lim_{x \to 0} \frac{\cos x - 1}{x} = 0$.
• Algebraic manipulation, combined with these theorems, can help solve a wide range of limit problems.