/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 83 The following limits equal the d... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 83

The following limits equal the derivative of a function \(f\) at a point \(a\) a. Find one possible \(f\) and \(a\) b. Evaluate the limit. $$\lim _{x \rightarrow \pi / 4} \frac{\cot x-1}{x-\frac{\pi}{4}}$$

Short Answer

Expert verified
In this problem, we found that the function \(f(x) = \tan(x)\) and the point \(a = \frac{\pi}{4}\) satisfy the given limit condition. The limit is equal to the derivative of the function at the specified point, and we have evaluated it as: $$\lim _{x \rightarrow \pi / 4} \frac{\cot x-1}{x-\frac{\pi}{4}} = f'\left(\frac{\pi}{4}\right) = \sec^2\left(\frac{\pi}{4}\right) = 2$$

Step by step solution

01

Finding f(x) and a

To find the function and the point where its derivative is equal to the given limit, we observe two key elements in the limit: the cotangent function and the point \(\frac{\pi}{4}\). We know that the derivative of the tangent function is \(\sec^2(x)\), so it's a good idea to check if the tangent function is our answer. Let \(f(x) = \tan(x)\). Now, we want to find the value of \(a\) such that the limit represents \(f'(a)\). Since the point in the limit is \(\frac{\pi}{4}\), let's set \(a = \frac{\pi}{4}\). Now, we need to find the derivative \(f'(x)\) and see if it matches the given limit.
02

Finding f'(x)

Given \(f(x) = \tan(x)\), we'll now calculate its derivative, \(f'(x)\), with respect to \(x\). The derivative of tangent function is: $$f'(x) = \sec^2(x)$$ Now, let's evaluate \(f'\left(\frac{\pi}{4}\right)\): $$f'\left(\frac{\pi}{4}\right) = \sec^2\left(\frac{\pi}{4}\right) = \left(\frac{1}{\cos\left(\frac{\pi}{4}\right)}\right)^2 = 2$$ So, we were looking for the point a and function f. Announcing the result is: $$f(x) = \tan(x), \quad a=\frac{\pi}{4}$$
03

Evaluating the given limit

The limit we have to evaluate is: $$\lim _{x \rightarrow \pi / 4} \frac{\cot x-1}{x-\frac{\pi}{4}}$$ We know that \(\cot(x) = \frac{1}{\tan(x)}\), so we can rewrite the limit as: $$\lim _{x \rightarrow \pi / 4} \frac{\frac{1}{\tan(x)}-1}{x-\frac{\pi}{4}}$$ Now we substitute our chosen function, \(f(x) = \tan{x}\), into the limit expression, which becomes: $$\lim _{x \rightarrow \pi / 4} \frac{ \frac{1}{f(x)}-1}{x-a}$$ Now, let's use the expression for the derivative of \(f(x)\) at point \(a\): $$f'(a) = \lim_{x \rightarrow a} \frac{f(x) - f(a)}{x-a} = \lim_{x \rightarrow a} \frac{f(x)}{x-a}$$ Multiplying numerator and denominator by \( f(x)f(a) \), we get $$f'(a) = \lim_{x \rightarrow a} \frac{f(x)f(a)-f(a)f(x)}{(x-a)f(x)f(a)}$$ which can be simplified to $$f'(a) = \lim_{x \rightarrow a} \frac{(1 - f(a))f(x)}{(x-a)f(x)f(a)}$$ Now, we have $$\lim _{x \rightarrow \pi / 4} \frac{\frac{1}{\tan(x)}-1}{x-\frac{\pi}{4}} = \lim_{x \rightarrow a} \frac{ \frac{1}{f(x)}-1}{x-a} = f'(a)$$ So, the given limit is equal to the derivative of \(f(x) = \tan(x)\) at point \(a = \frac{\pi}{4}\). Evaluating the limit, we get: $$\lim _{x \rightarrow \pi / 4} \frac{\cot x-1}{x-\frac{\pi}{4}} = f'\left(\frac{\pi}{4}\right) = 2$$

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Limit Definition of Derivative
The limit definition of the derivative is the fundamental concept that forms the basis of calculus. It enables us to compute the instantaneous rate of change of a function at a given point, which is the derivative of the function at that point. In more formal terms, the derivative of a function, denoted by \( f'(a) \), at a point \( a \) is defined by the limit:

\[ f'(a) = \lim_{{x \to a}} \frac{{f(x) - f(a)}}{{x - a}} \]
This expression states that to find the derivative of the function at \( a \), we compute the limit of the difference quotient as \( x \) approaches \( a \). The difference quotient measures the approximately change-per-unit of the function over an interval, and the limit process refines this to the exact rate of change at a single point.
Derivative of Tangent Function
When we discuss the derivative of the tangent function, we're considering how the tangent function changes at any given point on its curve. Given the function \( f(x) = \tan(x) \), its derivative is expressed as:

\[ f'(x) = \sec^2(x) \]
This equation implies that the rate of change of the tangent function with respect to \( x \) at any point is given by the square of the secant function of \( x \). The secant function, \( \sec(x) \), is simply the reciprocal of the cosine function, \( \cos(x) \). This derivative highlights unique trigonometric relationships and is particularly useful for solving problems involving angular velocities and related rates in physics.
Trigonometric Limits
The concept of trigonometric limits is essential when dealing with functions that involve trigonometry, especially as the function values approach specific points. Trigonometric limits help us understand the behavior of trigonometric functions near these points. For instance, common limits such as:

\[ \lim_{{x \to 0}} \frac{{\sin(x)}}{{x}} = 1 \]
or

\[ \lim_{{x \to 0}} \frac{{1 - \cos(x)}}{{x}} = 0 \]
are fundamental in calculus. Knowing these limits allows students to evaluate more complex limits that involve trigonometric functions. They often involve relationships that are not immediately obvious, but through the use of certain trigonometric identities and limit laws, these expressions can be simplified to find the desired limit.
Continuous Functions
A function is considered continuous at a point if there is no interruption in the graph of the function at that point. For a function to be continuous at point \( a \), three conditions must be satisfied:
  • The function \( f(a) \) must be defined.
  • The limit \( \lim_{{x \to a}} f(x) \) must exist.
  • The limit of the function as \( x \) approaches \( a \) must be equal to the function value at \( a \): \( \lim_{{x \to a}} f(x) = f(a) \).

If these conditions hold for every point in an interval, then the function is said to be continuous over that interval. A key property of continuous functions is that they can be graphed without lifting the pencil from the paper. Continuity is a crucial property for functions in calculus, and particularly important when applying the Intermediate Value Theorem, which states that a continuous function that takes on two different values will take on any intermediate value at some point within the domain.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An observer stands \(20 \mathrm{m}\) from the bottom of a 10 -m-tall Ferris wheel on a line that is perpendicular to the face of the Ferris wheel. The wheel revolves at a rate of \(\pi \mathrm{rad} / \mathrm{min},\) and the observer's line of sight with a specific seat on the wheel makes an angle \(\theta\) with the ground (see figure). Forty seconds after that seat leaves the lowest point on the wheel, what is the rate of change of \(\theta ?\) Assume the observer's eyes are level with the bottom of the wheel.

a. Determine an equation of the tangent line and normal line at the given point \(\left(x_{0}, y_{0}\right)\) on the following curves. b. Graph the tangent and normal lines on the given graph. $$\begin{aligned} &3 x^{3}+7 y^{3}=10 y\\\ &\left(x_{0}, y_{0}\right)=(1,1) \end{aligned}$$

Calculating limits exactly Use the definition of the derivative to evaluate the following limits. $$\lim _{h \rightarrow 0} \frac{(3+h)^{3+h}-27}{h}$$

Orthogonal trajectories Two curves are orthogonal to each other if their tangent lines are perpendicular at each point of intersection (recall that two lines are perpendicular to each other if their slopes are negative reciprocals). A family of curves forms orthogonal trajectories with another family of curves if each curve in one family is orthogonal to each curve in the other family. For example, the parabolas \(y=c x^{2}\) form orthogonal trajectories with the family of ellipses \(x^{2}+2 y^{2}=k,\) where \(c\) and \(k\) are constants (see figure). Find \(d y / d x\) for each equation of the following pairs. Use the derivatives to explain why the families of curves form orthogonal trajectories. \(y=m x ; x^{2}+y^{2}=a^{2},\) where \(m\) and \(a\) are constants

Tangency question It is easily verified that the graphs of \(y=1.1^{x}\) and \(y=x\) have two points of intersection, and the graphs of \(y=2^{x}\) and \(y=x\) have no points of intersection. It follows that for some real number \(1
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Probability and Statistics

Read Explanation

Mechanics Maths

Read Explanation

Applied Mathematics

Read Explanation

Calculus

Read Explanation

Discrete Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.