Problem 70 - Calculus Volume 1

Question

The given limit represents the derivative of a function \(y=f(x)\) at \(x=a\). Find \(f(x)\) and \(a\). $$ \lim _{h \rightarrow 0} \frac{\cos (\pi+h)+1}{h} $$

Step-by-Step Solution

Answer

The function is \( f(x) = \cos(x) + 1 \) and the point is \( a = \pi \).

Step 1: Recognize the Derivative Form

The given limit \( \lim _{h \rightarrow 0} \frac{\cos (\pi+h)+1}{h} \) is in the form \( \lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h} \). This represents the derivative of a function \( f(x) \) at a point \( x = a \). Our task is to identify \( f(x) \) and \( a \).

Step 2: Analyze Function Inside the Limit

Identify that the expression inside the limit, \( \cos(\pi+h) + 1 \), represents \( f(a+h) \). Notice that \( f(a) = \cos(\pi) + 1 \). Thus, the function \( f(x) \) can be rewritten as \( \cos(x) + 1 \).

Step 3: Identify Point a

From the expression \( \cos(\pi+h) + 1 \), we deduce that \( f(a+h) = \cos(a+h) + 1 \). Therefore, \( a = \pi \) because \( a + h \) corresponds to substituting \( x = \pi + h \).

Step 4: Verify the Function and Point

Verify that the function \( f(x) = \cos(x) + 1 \) and point \( a = \pi \) fit in the given derivative limit form. The expression \( \cos(\pi+h) + 1 \) indeed aligns with substituting this \( f(x) \) at \( x = \pi + h \).

Key Concepts

Understanding Limits

One of the key concepts in calculus is the limit. A limit helps us understand the behavior of functions as they approach a certain point or value. In our exercise, we have a limit expressed as \[\lim _{h \rightarrow 0} \frac{\cos (\pi+h)+1}{h}\]which is crucial in finding derivatives. When we say a limit evaluates the derivative, it means we're looking at how a function changes—it's the function's rate of change at a particular point. For the provided expression:

• The limit explores what happens to the value of the fraction as \( h \) gets infinitesimally close to 0.
• This form represents a fundamental concept called the derivative.
• Understanding this requires recognizing the pattern \( \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \), analogous to quantifying the slope of a tangent to the curve defined by the function.

A solid grasp of limits is imperative to calculus because it forms the foundation of derivatives and integrals. To master it, practice recognizing forms like the standard derivative form used in this exercise.

Trigonometric Functions: The Role of Cosine

Trigonometric functions, such as sine and cosine, are fundamental in mathematics, particularly in calculus. They allow us to model and analyze periodic phenomena, and know about their properties is essential for calculus problem solving.
In the given exercise, we have the expression \( \cos(\pi + h) + 1 \), which we identified to rewrite the derivative form. Let's break it down:

• \( \cos(\pi + h) \) uses the cosine of the sum formula, which is a trigonometric identity.
• The function \( f(x) = \cos(x) + 1 \) stems from understanding that by setting \( x = \pi + h \), it aligns with the problem's limit structure.

For any angle \( x \), cosine has specific properties:

• Cosine is even, meaning \( \cos(-x) = \cos(x) \).
• \( \cos(\pi) = -1 \), simplifying many problems, including showing \( f(a) = \cos(\pi) + 1 = 0 \).

Clearly understanding these functions enables pattern recognition and manipulation of expressions within calculus problems.

Principles of Calculus Problem Solving

Solving calculus problems often requires a systematic approach, involving identifying known patterns, applying rules, and verifying results. The limit given in this exercise represents such a methodical process. To solve the problem, you need to:

• Recognize patterns like the derivative form: \( \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \).
• Break down expressions considering their function properties—here, the trigonometric function \( f(x) = \cos(x) + 1 \).
• Link each part of the expression to the problem's context, identifying functions and points \( a \), helping to clear the solution logically.
• Verify and double-check the proposed values for function and point satisfy the original limit's requirements.

Practicing these steps with different calculus problems strengthens problem-solving techniques. Thoroughly understanding each part and how they work together ensures you can tackle similar problems with confidence in the future.