/*! 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 60 Evaluate the following limits or... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 60

Evaluate the following limits or explain why they do not exist. Check your results by graphing. $$\lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{\ln x}$$

Short Answer

Expert verified
Answer: The limit of the function as $x$ approaches infinity is 1.

Step by step solution

01

Rewrite the expression using exp and ln functions

We can rewrite the given expression using the property \(a^{\log_{a}{b}} = b\) as: $$\lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{\ln x} = \lim_{x \rightarrow \infty} e^{\ln((1+\frac{1}{x})^{\ln x})}$$
02

Simplify the exponent using the property of logarithms

Use the property of logarithms \(\ln(a^b) = b \cdot \ln(a)\) to simplify the exponent: $$\lim_{x \rightarrow \infty} e^{\ln((1+\frac{1}{x})^{\ln x})} = \lim_{x \rightarrow \infty} e^{\ln x \cdot \ln(1 + \frac{1}{x})}$$
03

Apply L'Hopital's Rule to the exponent

Now, we have a limit of the form \(e^{\frac{0}{0}}\), as both \(\ln x\) and \(\ln(1 + \frac{1}{x})\) approach 0 as x approaches infinity. In this situation, we can apply L'Hopital's Rule to the expression in the exponent to deal with the indeterminate form: $$\lim_{x \rightarrow \infty} e^{\frac{\ln x \cdot \ln(1 + \frac{1}{x})}{1}} = \lim_{x \rightarrow \infty} e^{\frac{\frac{1}{x}\cdot \frac{-(1/x^2)}{1+1/x}}=\lim_{x\rightarrow\infty}e^{\frac{-1/x}{1+1/x}}}$$
04

Evaluate the limit

We can now evaluate this limit as x approaches infinity: $$\lim_{x\rightarrow\infty}e^{\frac{-1/x}{1+1/x}} = e^{\frac{-1/\infty}{1+1/\infty}}=e^{\frac{0}{1}}$$
05

Final Result

The limit evaluates to $$e^{\frac{0}{1}} = e^0 = 1$$ Therefore, the limit of the given function as x approaches infinity is 1.

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

Consider the limit \(\lim _{x \rightarrow \infty} \frac{\sqrt{a x+b}}{\sqrt{c x+d}},\) where \(a, b, c\) and \(d\) are positive real numbers. Show that l'H么pital's Rule fails for this limit. Find the limit using another method.

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. a. The linear approximation to \(f(x)=x^{2}\) at \(x=0\) is \(L(x)=0\) b. Linear approximation at \(x=0\) provides a good approximation to \(f(x)=|x|\) c. If \(f(x)=m x+b,\) then the linear approximation to \(f\) at any point is \(L(x)=f(x)\)

Economists use demand functions to describe how much of a commodity can be sold at varying prices. For example, the demand function \(D(p)=500-10 p\) says that at a price of \(p=10,\) a quantity of \(D(10)=400\) units of the commodity can be sold. The elasticity \(E=\frac{d D}{d p} \frac{p}{D}\) of the demand gives the approximate percent change in the demand for every \(1 \%\) change in the price. a. Compute the elasticity of the demand function \(D(p)=500-10 p\) b. If the price is \(\$ 12\) and increases by \(4.5 \%,\) what is the approximate percent change in the demand? c. Show that for the linear demand function \(D(p)=a-b p\) where \(a\) and \(b\) are positive real numbers, the elasticity is a decreasing function, for \(p \geq 0\) and \(p \neq a / b\) d. Show that the demand function \(D(p)=a / p^{b}\), where \(a\) and \(b\) are positive real numbers, has a constant elasticity for all positive prices.

Given the following acceleration functions of an object moving along a line, find the position function with the given initial velocity and position. $$a(t)=2 e^{-t / 6} ; v(0)=1, s(0)=0$$

Suppose that object A is located at \(s=0\) at time \(t=0\) and starts moving along the \(s\) -axis with a velocity given by \(v(t)=2 a t,\) where \(a > 0 .\) Object \(B\) is located at \(s=c>0\) at \(t=0\) and starts moving along the \(s\) -axis with a constant velocity given by \(V(t)=b>0 .\) Show that \(\mathrm{A}\) always overtakes \(\mathrm{B}\) at time $$t=\frac{b+\sqrt{b^{2}+4 a c}}{2 a}$$.
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Mechanics Maths

Read Explanation

Calculus

Read Explanation

Geometry

Read Explanation

Pure Maths

Read Explanation

Applied 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.