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Chapter 3: Problem 66

(Continuation of Exercise \(65 .)\) The quotient $$ \frac{f(x+h)-f(x-h)}{2 h} $$ may have a limit as \(h \rightarrow 0\) when \(f\) has no derivative at \(x .\) As a case in point, take \(f(x)=|x|\) and calculate $$ \lim _{h \rightarrow 0} \frac{|0+h|-|0-h|}{2 h} $$ As you will see, the limit exists even though \(f(x)=|x|\) has no derivative at \(x=0 .\) Moral: Before using a centered difference quotient, be sure the derivative exists.

Short Answer

Expert verified
The limit \( \lim_{h \rightarrow 0} \frac{|0+h|-|0-h|}{2 h} \) exists and equals 0.

Step by step solution

01

Understand the Function

The function given is \( f(x) = |x| \). The absolute value function is not differentiable at \( x = 0 \) because the left-hand and right-hand derivatives do not match at this point.
02

Set Up the Limit Expression

We are tasked with calculating the limit \( \lim_{h \to 0} \frac{|0 + h| - |0 - h|}{2h} \). Substitute \( x = 0 \) into the expression \( \frac{f(x+h) - f(x-h)}{2h} \).
03

Simplify the Numerator

Substitute the expression: \( |0 + h| = |h| \) and \( |0 - h| = |-h| \). Both \(|h|\) and \(|-h|\) equal \(h\) when \(h > 0\) and \(-h\) when \(h < 0\). So, \(|0+h| - |0-h| = |h| - |-h| = 0\).
04

Evaluate the Limit

The expression simplifies to \( \frac{0}{2h} = 0 \). Therefore, the limit as \( h \to 0 \) is \(0\). Since there is no dependency on \( h \) after simplification, the limit exists and is equal to 0.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Absolute Value Function
The absolute value function, denoted as \( f(x) = |x| \), is a function that outputs the non-negative value of its input \( x \). This means if \( x \) is positive or zero, \( |x| = x \) and if \( x \) is negative, \( |x| = -x \). The function features a "V" shape when plotted on a graph because it reflects negative values to positive ones.

One interesting property of the absolute value function is its behavior at the center, \( x = 0 \). At this point, the function is continuous but not differentiable. The reason this happens is that the function sharply changes direction, leading to a discontinuity in the derivative. This is known as a cusp.

In the original problem, the absolute value function is explored to demonstrate its behavior at a sharp corner. By understanding these properties, students learn why both limits and differentiability are crucial concepts when analyzing functions involving absolute values.
Differentiability
Differentiability is a fascinating concept in calculus that describes a function's ability to have a derivative at a particular point. When we say a function is differentiable at a certain point, it means that there exists a well-defined tangent line to the function at that point.

For a function to be differentiable, it must be smooth, with no sharp turns or cusps, at that specific point. In the context of the absolute value function, \( f(x) = |x| \), it is not differentiable at \( x = 0 \) due to its "V" shape. At this cusp, the left-hand derivative (approaching from the negative side) and the right-hand derivative (approaching from the positive side) do not match.
  • If left-hand and right-hand derivatives are equal, the derivative at that point exists.
  • If they are different, the function is not differentiable at that point, as seen in the absolute value function at \( x = 0 \).
Understanding differentiability is essential when dealing with the centered difference quotient because it highlights the importance of checking whether a function can have a derivative before computing it.
Limits in Calculus
Limits form the backbone of calculus, providing a rigorous way to understand and analyze the behavior of functions as they approach certain points or infinity. Calculating limits, especially as a variable approaches zero, allows us to understand how a function behaves near a specific point.

In the exercise involving the centered difference quotient, a limit is used to evaluate the expression \( \lim_{h \to 0} \frac{|0+h| - |0-h|}{2h} \). This expression computes the average rate of change of the absolute value function across a small interval near \( x = 0 \).
  • The importance of limits in this context lies in understanding the function's local behavior.
  • A function can have limits resolved even if it is not differentiable at that point, as is discussed in the original exercise.
  • The limit in the centered difference quotient is shown to exist despite the cusp in the absolute value function because the expression reduces to zero.
Learning about limits strengthens the foundational knowledge needed to grasp advanced concepts in calculus, such as continuity and differentiability.

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Most popular questions from this chapter

When the length \(L\) of a clock pendulum is held constant by controlling its temperature, the pendulum's period \(T\) depends on the acceleration of gravity \(g\). The period will therefore vary slightly as the clock is moved from place to place on the earth's surface, depending on the change in \(g.\) By keeping track of \(\Delta T,\) we can estimate the variation in \(g\) from the equation \(T=2 \pi(L / g)^{1 / 2}\) that relates \(T, g,\) and \(L\). a. With \(L\) held constant and \(g\) as the independent variable, calculate \(d T\) and use it to answer parts (b) and (c). b. If \(g\) increases, will \(T\) increase or decrease? Will a pendulum clock speed up or slow down? Explain. c. A clock with a \(100-\mathrm{cm}\) pendulum is moved from a location where \(g=980 \mathrm{cm} / \mathrm{s}^{2}\) to a new location. This increases the period by \(d T=0.001\) s. Find \(d g\) and estimate the value of \(g\) at the new location.

Use a CAS to estimate the magnitude of the error in using the linearization in place of the function over a specified interval \(I\). Perform the following steps: a. Plot the function \(f\) over \(I\) b. Find the linearization \(L\) of the function at the point \(a\) c. Plot \(f\) and \(L\) together on a single graph. d. Plot the absolute error \(|f(x)-L(x)|\) over \(I\) and find its maximum value. e. From your graph in part (d), estimate as large a \(\delta>0\) as you can, satisfying \(|x-a|<\delta \quad \Rightarrow \quad|f(x)-L(x)|<\epsilon\) for \(\epsilon=0.5,0.1,\) and \(0.01 .\) Then check graphically to see if your \(\delta\) -estimate holds true. $$f(x)=\sqrt{x}-\sin x, \quad[0,2 \pi], \quad a=2$$

Find the value of \((f \circ g)^{\prime}\) at the given value of \(x\). $$f(u)=\frac{2 u}{u^{2}+1}, \quad u=g(x)=10 x^{2}+x+1, \quad x=0$$

Use a CAS to estimate the magnitude of the error in using the linearization in place of the function over a specified interval \(I\). Perform the following steps: a. Plot the function \(f\) over \(I\) b. Find the linearization \(L\) of the function at the point \(a\) c. Plot \(f\) and \(L\) together on a single graph. d. Plot the absolute error \(|f(x)-L(x)|\) over \(I\) and find its maximum value. e. From your graph in part (d), estimate as large a \(\delta>0\) as you can, satisfying \(|x-a|<\delta \quad \Rightarrow \quad|f(x)-L(x)|<\epsilon\) for \(\epsilon=0.5,0.1,\) and \(0.01 .\) Then check graphically to see if your \(\delta\) -estimate holds true. $$f(x)=x^{3}+x^{2}-2 x, \quad[-1,2], \quad a=1$$

The line that is normal to the curve \(x^{2}+2 x y-3 y^{2}=0\) at (1,1) intersects the curve at what other point?
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