Problem 21 - Calculus

Question

Show that \((\tan h) / h\) is squeezed between 1 and \(1 / \cos h .\) As \(h \rightarrow 0\) the limit is ______ .

Step-by-Step Solution

Answer

1.

Step 1: Identify Known Limit Properties

Introduce the limit properties for trigonometric functions. As \( h \rightarrow 0 \), \( \sin h \rightarrow h\) and \( \cos h \rightarrow 1 \). Therefore, \( \tan h \rightarrow h\).

Step 2: Rewrite the Expression

Express \( \frac{\tan h}{h} = \frac{\sin h}{h} \cdot \frac{1}{\cos h} \). Here, \( \frac{\sin h}{h} \) tends to 1 as \( h \rightarrow 0 \), and requires additional examination for \( \frac{1}{\cos h} \).

Step 3: Examine \( \frac{1}{\cos h} \)

Since \( \cos h \rightarrow 1 \) as \( h \rightarrow 0 \), \( \frac{1}{\cos h} \rightarrow \frac{1}{1} = 1\). This implies that as \( h \rightarrow 0 \), \( \frac{1}{\cos h} \) approaches 1.

Step 4: Establish Inequalities

Show the inequality \( 1 \leq \frac{\sin h}{h} \leq \frac{1}{\cos h} \). Since we have \( \frac{\tan h}{h} = \frac{\sin h}{h} \cdot \frac{1}{\cos h} \), we can multiply both sides of the inequality \( 1 \leq \frac{\sin h}{h} \leq \frac{1}{\cos h} \) by \( \frac{1}{\cos h} \), maintaining the inequality since \( \frac{1}{\cos h} \geq 1 \). We derive:\[ 1 \leq \frac{\tan h}{h} \leq \frac{1}{\cos h} \].

Step 5: Conclude by Squeeze Theorem

Using the Squeeze Theorem, since \( \,\frac{\tan h}{h}\, \) is squeezed between \( 1 \,\) and \( \,\frac{1}{\cos h} \to 1 \,\) as \( \,h \rightarrow 0\), the limit of \( \,\frac{\tan h}{h}\,\) is 1.

Key Concepts

Trigonometric Limits

When working with trigonometric limits, especially as a variable approaches zero, certain patterns and behaviors emerge. Consider the limit of \( \sin h \) and \( \cos h \) as \( h \rightarrow 0 \). Here, \( \sin h \rightarrow h \) and \( \cos h \rightarrow 1 \). This reflects the small-angle approximations, where \( \sin h \approx h \) for small angles, making calculations more straightforward. It helps us understand why \( \tan h \approx h \) in this particular limit.

The expression \( \frac{\tan h}{h} \) can further be broken down using trigonometric identities into \( \frac{\sin h}{h} \cdot \frac{1}{\cos h} \). Here, \( \frac{\sin h}{h} \rightarrow 1 \) as \( h \rightarrow 0 \), owing to the nature of sine and its behavior around \( 0 \). Meanwhile, \( \frac{1}{\cos h} \rightarrow 1 \) as well because the cosine of a very small angle approaches 1.

These foundational behaviors are essential when proving the limits involving trigonometric functions and play a critical role in solving problems like the one presented. Understanding these helps decode more complex expressions and eases calculus computations.

Limit Properties

The cornerstone of calculus, limit properties, allows us to understand how functions behave as they approach a particular point. This is vital when dealing with complicated trigonometric functions.

• Product and Quotient Rules: These properties indicate how the limits of product or quotient of functions can be found and used jointly, as in \( \frac{\tan h}{h} \).
• Standard Limits: Familiar limits such as \( \lim_{{h \to 0}} \frac{\sin h}{h} = 1 \) and \( \lim_{{h \to 0}} \frac{1}{\cos h} = 1 \) simplify the process.

Breaking \( \frac{\tan h}{h} \) into more manageable parts, namely \( \frac{\sin h}{h} \) and \( \frac{1}{\cos h} \), demonstrates these properties at work. When each component's limit behavior is known, solving the overarching problem becomes more systematic. This specific limit scenario exploits these properties to validate inequalities and ultimately apply the Squeeze Theorem effectively.

Inequality Proof

Inequality proofs often involve demonstrating that one expression is squeezed or bounded between two others. In this exercise, the key is to show that \( 1 \leq \frac{\sin h}{h} \leq \frac{1}{\cos h} \).

Once established, these inequalities work as follows:

• Using the fact that \( \frac{\cos h}{h} \) tends to 1 as \( h \rightarrow 0 \), multiply each part of the inequality \( 1 \leq \frac{\sin h}{h} \leq \frac{1}{\cos h} \) by \( \frac{1}{\cos h} \).
• The calculated inequality: \( 1 \leq \frac{\tan h}{h} \leq \frac{1}{\cos h} \) shows that as \( h \rightarrow 0 \), both upper and lower bounds converge to 1.

The Squeeze Theorem comes into play to conclude the proof. Since the `\( \frac{\tan h}{h} \)` expression is bounded by two expressions that approach 1, by this theorem, its limit must also be 1. Thus, inequality proof not only lays a foundation for applying the Squeeze Theorem but also confirms the consistency of the function's behavior near the point of interest.