Problem 35 - Calculus Early Transcendentals: Pearson New International Edition

Question

\(\lim _{h \rightarrow 0} \frac{\cos (x+h)-\cos x}{h}\)

Step-by-Step Solution

Answer

The limit is \(-\sin x\).

Step 1: Recall Trigonometric Identity

To solve the limit, we will use the trigonometric identity for the difference of cosines: \[\cos(x+h) = \cos x \cdot \cos h - \sin x \cdot \sin h\] This identity will help us in simplifying the expression inside the limit.

Step 2: Apply the Identity

Substitute the trigonometric identity into the limit expression, replacing \(\cos(x+h)\): \[\lim_{h \to 0} \frac{\cos(x+h) - \cos x}{h} = \lim_{h \to 0} \frac{(\cos x \cdot \cos h - \sin x \cdot \sin h) - \cos x}{h}\] Simplify this to: \[\lim_{h \to 0} \frac{\cos x (\cos h - 1) - \sin x \sin h}{h}\]

Step 3: Split the Limit

We can separate the limit into two parts by applying the linearity property of limits: \[\lim_{h \to 0} \frac{\cos x (\cos h - 1) - \sin x \sin h}{h} = \cos x \lim_{h \to 0} \frac{\cos h - 1}{h} - \sin x \lim_{h \to 0} \frac{\sin h}{h}\]

Step 4: Evaluate Each Limit Separately

Now, solve each of the limits separately:1. \(\lim_{h \to 0} \frac{\cos h - 1}{h} = 0\) because \(\cos h\) approaches 1 much faster than \(h\) approaches 0.2. \(\lim_{h \to 0} \frac{\sin h}{h} = 1\) because it's a well-known trigonometric limit.

Step 5: Substitute and Simplify

Once evaluated, substitute the limit results back into the expression:\[\cos x \cdot 0 - \sin x \cdot 1 = -\sin x\] Thus, the limit simplifies to \(-\sin x\).

Key Concepts

Trigonometric Identities

Understanding trigonometric identities is essential when working with limits involving trigonometric functions. These identities are equations that hold true for all values within their domain, making them powerful tools for simplifying expressions.

In the context of limits, the difference of cosines identity is particularly useful. This identity states:

• \[\cos(x + h) = \cos x \cdot \cos h - \sin x \cdot \sin h\]

This allows us to rewrite expressions that include terms like \(\cos(x+h)\) into a form that can be more easily manipulated. In the solution above, this identity facilitates the simplification of the expression under the limit sign, helping you to evaluate the limit more straightforwardly.

Difference Quotient

The difference quotient is central to understanding derivatives, an integral part of calculus. It represents the average rate of change of a function over a small interval and is expressed as:

• \[\frac{f(x+h) - f(x)}{h}\]

In terms of limits, the goal is often to evaluate this expression as \(h\) approaches zero. The original exercise uses the difference quotient in the form of trigonometric functions, specifically using the cosine function. By understanding the behavior of the trigonometric function in this quotient, you can gain insights into the derivative of the function as \(h\) approaches zero.

Limit of Trigonometric Functions

Evaluating the limit of trigonometric functions is a common task in calculus. These limits can sometimes be counterintuitive, so it helps to know some basic limits. For example, a critical limit to remember is:

• \[\lim_{h \to 0} \frac{\sin h}{h} = 1\]
• \[\lim_{h \to 0} \frac{\cos h - 1}{h} = 0\]

These are fundamental limits frequently encountered when dealing with sine and cosine functions. In the exercise, applying these known limits allows us to break and compute the simplified expressions during the evaluation process. Using these known results simplifies complex expressions and aids in finding the final answer.

Calculus Middle Level

At the calculus middle level, students dive deeper into concepts of differentiation and the analysis of functions. Solving limits of trigonometric functions is a stepping stone to understanding derivatives. It provides practice in:

• Applying trigonometric identities to simplify complex functions.
• Utilizing limit properties effectively to separate and evaluate different parts of an expression.

In this exercise, you're getting practice with both, using identities to simplify the given expression and employing limit properties to isolate and solve for component limits individually. Successfully applying these techniques is crucial for mastering more advanced topics in calculus later on.