/*! 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 14 Find the limit of the following ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the limit of the following sequences or determine that the limit does not exist. $$\left\\{\frac{k}{\sqrt{9 k^{2}+1}}\right\\}$$

Short Answer

Expert verified
Answer: The limit of the sequence exists and is equal to 0.

Step by step solution

01

Identify the largest exponent of k

In the given sequence, the largest exponent of k is 2 (found in the expression \(9k^2\))
02

Divide both numerator and denominator by the largest exponent of k

We will divide both the numerator and the denominator of the expression by \(k^2\): $$\frac{k}{\sqrt{9 k^{2}+1}}\times \frac{\frac{1}{k}}{\frac{1}{k}} = \frac{\frac{k}{k^2}}{\sqrt{\frac{9 k^{2}}{k^2}+\frac{1}{k^2}}} = \frac{\frac{1}{k}}{\sqrt{9+ \frac{1}{k^2}}}$$
03

Evaluate the limit

Now, we will evaluate the limit of the new expression as k approaches infinity: $$\lim_{k \to \infty} \frac{\frac{1}{k}}{\sqrt{9+ \frac{1}{k^2}}}$$ As k approaches infinity, \(\frac{1}{k}\) and \(\frac{1}{k^2}\) approach 0. So, the expression becomes: $$\lim_{k\to\infty}\frac{0}{\sqrt{9+0}} = \frac{0}{\sqrt{9}} = \frac{0}{3}$$ Thus, the limit of the given sequence exists and is equal to 0.

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 91Ó°ÊÓ!

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 following situations that generate a sequence. a. Write out the first five terms of the sequence. b. Find an explicit formula for the terms of the sequence. c. Find a recurrence relation that generates the sequence. d. Using a calculator or a graphing utility, estimate the limit of the sequence or state that it does not exist. The Consumer Price Index (the CPI is a measure of the U.S. cost of living) is given a base value of 100 in the year \(1984 .\) Assume the CPI has increased by an average of \(3 \%\) per year since \(1984 .\) Let \(c_{n}\) be the CPI \(n\) years after \(1984,\) where \(c_{0}=100.\)

An early limit Working in the early 1600 s, the mathematicians Wallis, Pascal, and Fermat were attempting to determine the area of the region under the curve \(y=x^{p}\) between \(x=0\) and \(x=1\) where \(p\) is a positive integer. Using arguments that predated the Fundamental Theorem of Calculus, they were able to prove that $$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=0}^{n-1}\left(\frac{k}{n}\right)^{p}=\frac{1}{p+1}$$ Use what you know about Riemann sums and integrals to verify this limit.

Showing that \(\sum_{k=1}^{\infty} \frac{1}{k^{2}}=\frac{\pi^{2}}{6} \operatorname{In} 1734,\) Leonhard Euler informally proved that \(\sum_{k=1}^{\infty} \frac{1}{k^{2}}=\frac{\pi^{2}}{6} .\) An elegant proof is outlined here that uses the inequality $$ \cot ^{2} x<\frac{1}{x^{2}}<1+\cot ^{2} x\left(\text { provided that } 0

A glimpse ahead to power series Use the Ratio Test to determine the values of \(x \geq 0\) for which each series converges. $$\sum_{k=1}^{\infty} \frac{x^{k}}{2^{k}}$$

Use Exercise 89 to determine how many terms of each series are needed so that the partial sum is within \(10^{-6}\) of the value of the series (that is, to ensure \(R_{n}<10^{-6}\) ). a. \(\sum_{k=0}^{\infty} 0.6^{k}\) b. \(\sum_{k=0}^{\infty} 0.15^{k}\)

See all solutions

Recommended explanations on Math Textbooks

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.