/*! 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 102 Calculating limits exactly Use t... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 102

Calculating limits exactly Use the definition of the derivative to evaluate the following limits. $$\lim _{h \rightarrow 0} \frac{\ln \left(e^{8}+h\right)-8}{h}$$

Short Answer

Expert verified
Answer: The value of the limit is $$\frac{1}{e^8}$$.

Step by step solution

01

Identify the function

In this case, the function is given by the natural logarithm, i.e., $$f(x) = \ln(x)$$. The limit is set up like the definition of the derivative, with the function being evaluated at $$e^8$$, an expression of the form $$f(a + h) - f(a)$$.
02

Simplify the given limit using logarithmic properties

We can simplify the given limit using the logarithmic property of the natural logarithm, which states that $$\ln(e^k) = k$$ for any $$k$$: $$\lim _{h \rightarrow 0} \frac{\ln \left(e^{8}+h\right)-8}{h}$$
03

Apply the derivative definition to the natural logarithm

The definition of the derivative with respect to $$x$$ of the function $$f(x) = \ln(x)$$ is as follows: $$f'(x) = \lim_{h \rightarrow 0} \frac{\ln(x + h) - \ln(x)}{h}$$ In this case, $$x = e^8$$, so the limit we need to compute is: $$\lim _{h \rightarrow 0} \frac{\ln \left(e^{8}+h\right)-\ln(e^8)}{h}$$
04

Use the logarithmic property of difference

We can further simplify using the logarithmic property that states $$\ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right)$$: $$\lim _{h \rightarrow 0} \frac{\ln \left(\frac{e^{8}+h}{e^8}\right)}{h}$$
05

Simplify and apply limit properties

Now, we can factor out a $$\frac{1}{e^8}$$ from the argument and simplify: $$\lim _{h \rightarrow 0} \frac{\ln \left(1+\frac{h}{e^8}\right)}{h}$$ We now have a limit of the form $$\lim_{x \rightarrow 0} \frac{\ln(1+x)}{x}$$. Let $$x = \frac{h}{e^8}$$, then we have: $$\lim _{x \rightarrow 0} \frac{\ln \left(1+x\right)}{x}$$
06

Evaluate the limit

The limit we have now is a well-known limit in calculus, which is equal to $$1$$: $$\lim _{x \rightarrow 0} \frac{\ln(1+x)}{x} = 1$$
07

Compute the final answer

Since we had made a substitution of $$x = \frac{h}{e^8}$$, we need to account for this to find the original limit. To do this, we simply multiply the result by the coefficient of $$h$$ in our substitution, which is $$\frac{1}{e^8}$$: $$\lim _{h \rightarrow 0} \frac{\ln \left(e^{8}+h\right)-8}{h} = \frac{1}{e^8}$$ Therefore, the final answer is: $$\lim _{h \rightarrow 0} \frac{\ln \left(e^{8}+h\right)-8}{h} = \frac{1}{e^8}$$

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!

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

The magnitude of the gravitational force between two objects of mass \(M\) and \(m\) is given by \(F(x)=-\frac{G M m}{x^{2}}\) where \(x\) is the distance between the centers of mass of the objects and \(G=6.7 \times 10^{-11} \mathrm{N}-\mathrm{m}^{2} / \mathrm{kg}^{2}\) is the gravitational constant (N stands for newton, the unit of force; the negative sign indicates an attractive force). a. Find the instantaneous rate of change of the force with respect to the distance between the objects. b. For two identical objects of mass \(M=m=0.1 \mathrm{kg},\) what is the instantaneous rate of change of the force at a separation of \(x=0.01 \mathrm{m} ?\) c. Does the instantaneous rate of change of the force increase or decrease with the separation? Explain.

The population of a culture of cells after \(t\) days is approximated by the function \(P(t)=\frac{1600}{1+7 e^{-0.02 t}},\) for \(t \geq 0\). a. Graph the population function. b. What is the average growth rate during the first 10 days? c. Looking at the graph, when does the growth rate appear to be a maximum? d. Differentiate the population function to determine the growth rate function \(P^{\prime}(t)\). e. Graph the growth rate. When is it a maximum and what is the population at the time that the growth rate is a maximum?

A cylindrical tank is full at time \(t=0\) when a valve in the bottom of the tank is opened. By Torricelli's Law, the volume of water in the tank after \(t\) hours is \(V=100(200-t)^{2}\), measured in cubic meters. a. Graph the volume function. What is the volume of water in the tank before the valve is opened? b. How long does it take the tank to empty? c. Find the rate at which water flows from the tank and plot the flow rate function. d. At what time is the magnitude of the flow rate a minimum? A maximum?

Prove the following identities and give the values of \(x\) for which they are true. $$\tan \left(2 \tan ^{-1} x\right)=\frac{2 x}{1-x^{2}}$$

Work carefully Proceed with caution when using implicit differentiation to find points at which a curve has a specified slope. For the following curves, find the points on the curve (if they exist) at which the tangent line is horizontal or vertical. Once you have found possible points, make sure they actually lie on the curve. Confirm your results with a graph. $$x\left(1-y^{2}\right)+y^{3}=0$$
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Probability and Statistics

Read Explanation

Statistics

Read Explanation

Calculus

Read Explanation

Geometry

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.