Question

A function \(f\) and \(a\) value \(a\) are given. Approximate the limit of the difference quotient, \(\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h},\) using \(h=\pm 0.1, \pm 0.01 .\) $$ f(x)=\sin x, \quad a=\pi $$

Short answer

The limit is approximately -1.

Step-by-step solution

Understand the Problem

The problem asks us to approximate the derivative of the function \( f(x) = \sin x \) at \( x = \pi \) by calculating the limit of the difference quotient as \( h \) approaches 0. The given values for \( h \) are \( 0.1, -0.1, 0.01, \) and \( -0.01 \).

Set Up the Difference Quotient

The difference quotient is given by \( \frac{f(a+h) - f(a)}{h} \). Plugging in the function \( f(x) = \sin x \) and \( a = \pi \), it becomes \( \frac{\sin(\pi + h) - \sin(\pi)}{h} \). Since \( \sin(\pi) = 0 \), this simplifies to \( \frac{\sin(\pi + h)}{h} \).

Evaluate for Different h Values

Calculate the difference quotient for each given \( h \):- For \( h = 0.1 \), the difference quotient is \( \frac{\sin(\pi + 0.1)}{0.1} \).- For \( h = -0.1 \), the difference quotient is \( \frac{\sin(\pi - 0.1)}{-0.1} \).- For \( h = 0.01 \), the difference quotient is \( \frac{\sin(\pi + 0.01)}{0.01} \).- For \( h = -0.01 \), the difference quotient is \( \frac{\sin(\pi - 0.01)}{-0.01} \).

Plug Values Into Calculator

Use a calculator to find the approximate values:- \( \sin(\pi + 0.1) \approx \sin(3.241) \approx -0.0998 \), so the difference quotient is \( \frac{-0.0998}{0.1} = -0.998 \).- \( \sin(\pi - 0.1) \approx \sin(3.041) \approx 0.0998 \), so the difference quotient is \( \frac{0.0998}{-0.1} = -0.998 \).- \( \sin(\pi + 0.01) \approx \sin(3.15159265) \approx -0.00999983 \), so the difference quotient is \( \frac{-0.00999983}{0.01} = -0.999983 \).- \( \sin(\pi - 0.01) \approx \sin(3.13159265) \approx 0.00999983 \), so the difference quotient is \( \frac{0.00999983}{-0.01} = -0.999983 \).

Conclude with the Approximation

The approximations for the limit of the difference quotient using the specified \( h \) values are all close to -1. Thus, we approximate \( \lim_{h \to 0} \frac{\sin(\pi + h) - \sin(\pi)}{h} \approx -1 \), which corresponds with the derivative of \( \sin x \) at \( x = \pi \), \( \cos(\pi) = -1 \).

Key concepts

Understanding Derivative Approximation

Derivative approximation is a way to estimate the slope of a curve at a given point. It's like finding the tangent line to the curve right there. This is vital in calculus because it helps in understanding how a function behaves locally. In our exercise, we're using the difference quotient to approximate the derivative of the function \( f(x) = \sin(x) \) at \( x = \pi \).
An approximation is necessary as it may not always be straightforward to compute derivatives directly. The difference quotient is a fraction that measures how much \( f(x) \) changes as \( x \) does. It's calculated with:

• \( \frac{f(a+h) - f(a)}{h} \) where \( a \) is the point of interest.
• \

Exploring Trigonometric Function Limits

Trigonometric function limits are crucial when dealing with the behaviors of sine, cosine, and tangent around specific points. These limits help us understand what happens to the function values as they approach certain angles.
In this exercise, the focus is on how the sine function behaves around the angles near \( \pi \). The sine of these angles differs slightly from zero. As \( h \) approaches zero, these small deviations allow us to estimate the slope or derivative.

Remember the special angle properties:

• The limit as \( h \) approaches zero of \( \frac{\sin(h)}{h} = 1 \).
• \( \sin(\pi) = 0 \) is exactly zero, hence the interest in how it behaves when \( h \) deviates a little from zero, thus examining \( \sin(\pi + h) \).

Understanding these limits is key in calculating the difference quotient effectively.

Unveiling the Sine Function

The sine function, \( \sin(x) \), is a fundamental trigonometric function that measures the height of a point on the unit circle corresponding to an angle \( x \). It's periodic, with a cycle every \( 2\pi \).
Its behavior around \( \pi \) is particularly interesting because the sine value there is 0. This means it's a point where the graph transitions smoothly, changing direction.

An important characteristic of \( \sin(x) \) when studying calculus is its relation to the derivative, \( \cos(x) \). At \( \pi \), as seen in this exercise, the derivative which is \( \cos(\pi) = -1 \), indicates a slope downwards right at the peak.

• Remember, \( \sin(x) \) relates directly to angles in a right triangle, also guiding changes with respect to the unit circle.
• Understanding this function deeply impacts your grasp of wave behaviours and oscillations.

Mastering Calculus Problem Solving

Calculus problem solving involves a series of logical steps to find solutions for limits, derivatives, and integrals. The method used reflects in solving real-world problems where dynamic change occurs.
In the exercise given, each differential steps aimed at simplifying and calculating the limit accurately. Such approach bridges pure mathematics with practical application.

To become proficient in calculus problem solving:

• Start by breaking down the problem: What are you trying to find? Here, it was the limit of the difference quotient.
• Set up the correct expressions, as done with the quotient \( \frac{\sin(\pi + h)}{h} \).
• Evaluate systematically for different known values to get a broad perspective of the behavior, then conclude as done by approximating the derivative to be \( -1 \).

This structured approach makes solving complex calculus problems more manageable.