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 \(h=\pm 0.1, \pm 0.01 .\) $$ f(x)=\cos x, \quad a=\pi $$

Short answer

The derivative at \( a = \pi \) is 0.

Step-by-step solution

Define the Difference Quotient

The difference quotient is defined as \( \frac{f(a+h)-f(a)}{h} \) where \( a \) is the point of interest (\( \pi \) in this case) and \( h \) is a small increment. We'll compute this for \( h = \pm 0.1 \) and \( h = \pm 0.01 \).

Calculate f(a)

Substitute \( a = \pi \) into the function \( f(x) = \cos x \). Therefore, \( f(a) = \cos(\pi) = -1 \).

Compute the Difference Quotient for h = 0.1

Substitute \( a = \pi \) and \( h = 0.1 \) into the difference quotient: \[\frac{f(\pi + 0.1) - (-1)}{0.1} = \frac{\cos(\pi+0.1) + 1}{0.1}.\]Calculate \( \cos(\pi+0.1) \approx -0.995 \), then compute the expression to get a value of approximately \( 0.05 \).

Compute the Difference Quotient for h = -0.1

Substitute \( a = \pi \) and \( h = -0.1 \) into the difference quotient: \[\frac{f(\pi - 0.1) - (-1)}{-0.1} = \frac{\cos(\pi-0.1) + 1}{-0.1}.\]Calculate \( \cos(\pi-0.1) \approx -0.995 \), then compute the expression to get a value of approximately \( 0.05 \).

Compute the Difference Quotient for h = 0.01

Substitute \( a = \pi \) and \( h = 0.01 \) into the difference quotient: \[\frac{f(\pi + 0.01) - (-1)}{0.01} = \frac{\cos(\pi+0.01) + 1}{0.01}.\]Calculate \( \cos(\pi+0.01) \approx -0.99995 \), then compute the expression to get a value of approximately \( 0.005 \).

Compute the Difference Quotient for h = -0.01

Substitute \( a = \pi \) and \( h = -0.01 \) into the difference quotient: \[\frac{f(\pi - 0.01) - (-1)}{-0.01} = \frac{\cos(\pi-0.01) + 1}{-0.01}.\]Calculate \( \cos(\pi-0.01) \approx -0.99995 \), then compute the expression to get a value of approximately \( 0.005 \).

Average the Approximations

The average of the difference quotients for \( h = \pm 0.1 \) and \( \pm 0.01 \) can be calculated. The values obtained are very close to zero, which suggests that as \( h \) approaches zero, the limit is zero.

Conclusion

The limit of the difference quotient \( \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \) as computed with decreasing \( h \) values approaches 0. This indicates that the derivative of \( f(x) = \cos(x) \) at \( a = \pi \) is 0.

Key concepts

Limit

The concept of a limit is fundamental in calculus. It describes the value that a function approaches as the variable approaches a specific point. Understanding limits is crucial when we approximate derivatives using the difference quotient. In this case, we are examining the behavior of the ratio \( \frac{f(a+h)-f(a)}{h} \) as \( h \) gets smaller.

This process helps us understand how the function changes at a point, which is a stepping stone to finding derivatives. By analyzing limits, we can see what the value of the difference quotient approaches as \( h \) tends toward zero. Calculating this gives us a precise glimpse into the slope of the tangent line at any given point on a curve. For the cosine function here, this limit helps us find the derivative at \( a = \pi \).

Approximation

Approximation in calculus is a method of finding a value that is close enough to the correct answer, often within a specified tolerance. When dealing with limits and derivatives, we use approximation to help us handle functions that may not be easy to compute directly.

In our exercise, we used several small values of \( h \) (like \( \pm 0.1 \) and \( \pm 0.01 \)) to approximate the limit. This is a way to get closer to the actual change in the function. The smaller the \( h \), the better our approximation will be. By averaging the results from our approximations, we can predict what the function's behavior will be as \( h \) goes to zero. Essentially, approximation is our tool for inching closer to the exact value in a manageable way.

Derivative Calculation

Derivative calculation involves finding the derivative of a function, which tells us the rate at which the function is changing at any given point. This is the core idea behind using the difference quotient.

The critical part of derivative calculation is understanding how the difference quotient represents changes over finite intervals. By shrinking these intervals to an infinitely small size (approaching zero), we find the derivative. In our given exercise, calculating the difference quotient for decreases in \( h \) revealed that the limit approaches 0. This means at \( x = \pi \), the change in the cosine function's slope is nonexistent -- hence, the derivative is 0 at that point. This understanding is vital in calculus for interpreting and predicting the behavior of functions.

Trigonometric Function

Trigonometric functions such as sine and cosine play pivotal roles in calculus due to their periodic nature and applications in modeling waves and oscillations. In this exercise, we're working with the cosine function.

The cosine function, \( f(x) = \cos(x) \), is characterized by its wave-like pattern and a range of outputs from -1 to 1. Evaluating this at specific points, like \( a = \pi \), where the cosine value is -1, is a classic exercise in trigonometry.

Using trigonometric identities, such as \( \cos(\pi + x) = -\cos(x) \), helps us compute the required values for the difference quotient. These functions are foundational in numerous fields, including physics and engineering, due to their ability to describe periodic phenomena analytically and predictably.