/*! 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 23 Find the limit of the following ... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the limit of the following sequences or determine that the limit does not exist. $$\left\\{\frac{n}{e^{n}+3 n}\right\\}$$

Short Answer

Expert verified
Sequence: \(\frac{n}{e^{n} + 3n}\) Answer: The limit of the sequence as n approaches infinity is 0.

Step by step solution

01

Write the given sequence as a continuous function

Consider the given sequence as a continuous function of \(n\): $$f(n) = \frac{n}{e^{n}+3n}$$ We will find the limit of this function as n approaches infinity.
02

Identify the indeterminate form

To find the limit, evaluate the function as n approaches infinity: $$\lim_{n\to \infty} \frac{n}{e^{n}+3n}$$ Both numerator and denominator approach infinity, so we have the indeterminate form \(\frac{\infty}{\infty}\). We can apply L'Hopital's Rule to further evaluate the limit.
03

Apply L'Hopital's Rule

Apply L'Hopital's Rule by taking the derivative of the numerator and denominator with respect to n: $$\lim_{n\to \infty} \frac{\frac{d}{dn} (n)}{\frac{d}{dn} (e^{n}+3n)}$$ Differentiate the numerator and denominator: $$\lim_{n\to \infty} \frac{1}{e^{n}+3}$$
04

Evaluate the limit

As \(n\) approaches infinity, the term \(e^{n}\) also approaches infinity, making the denominator approach an infinite value. Thus, the limit of the given function becomes: $$\lim_{n\to \infty} \frac{1}{e^{n}+3} = 0$$ So the limit of the sequence is 0.

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Key Concepts

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

L'Hopital's Rule
L'Hopital's Rule is an incredibly useful method for dealing with indeterminate forms of limits like \( \frac{\infty}{\infty} \) or \( \frac{0}{0} \). When faced with these forms, we can simplify the limit by taking the derivative of the numerator and the derivative of the denominator separately.

Here’s how it works:
  • Identify if the function limit is in an indeterminate form.
  • Take the derivative of the top function and the derivative of the bottom function.
  • Re-evaluate the limit using the new function.
    For example, in the given exercise, the limit \( \lim_{n \to \infty} \frac{n}{e^n + 3n} \) was in the \( \frac{\infty}{\infty} \) form. By applying L'Hopital's Rule, we differentiated to get \( \lim_{n \to \infty} \frac{1}{e^n + 3} \).
This method can be used iteratively if the resulting function after applying L'Hopital's Rule is still an indeterminate form. It's important to recheck the form after each differentiation!

Remember, L'Hopital's Rule applies only to certain types of indeterminate forms. It's not a catch-all solution, but it's a powerful tool when applicable.
Limits of Functions
Understanding the limit of a function is a fundamental concept in calculus. It helps us make sense of how a function behaves as it approaches a certain point—often, infinity or some specific number.

A limit tells you the value that a function approaches as the input approaches some value. In our exercise, we're interested in what happens to the sequence's function \( f(n) = \frac{n}{e^n + 3n} \) as \( n \) tends to infinity.

The procedure typically involves:
  • Direct substitution to see if we get an indeterminate form.
  • If so, simplifying the function, perhaps with tools like L'Hopital's Rule, to find the limit value.
We concluded that \( \lim_{n \to \infty} \frac{n}{e^n + 3n} = 0 \) because, although both numerator and denominator grow without bound, the exponential function \( e^n \) grows much faster than \( n \) or \( 3n \), leading to the fraction approaching zero as \( n \) grows large.
Indeterminate Forms
Indeterminate forms occur when evaluating limits results in uncertain expressions like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). These forms do not directly tell us the value of the limit, and we must find ways to resolve them.

Common indeterminate forms include:
  • \( \frac{0}{0} \)
  • \( \frac{\infty}{\infty} \)
  • \( 0 \cdot \infty \)
  • \( \infty - \infty \)
  • \( 0^0 \)
  • \( 1^\infty \)
  • \( \infty^0 \)
In our problem, we started with the \( \frac{\infty}{\infty} \) form, typical when both the numerator and denominator of a sequence or function grow endlessly large.

To solve this, techniques like L'Hopital's Rule, algebraic manipulation, or applying limits that simplify the function can be used. Each situation might require a different approach, but once recognized, resolving indeterminate forms is a crucial skill for calculating limits accurately. With practice, you’ll become more comfortable recognizing and resolving these forms to evaluate limits properly.

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Most popular questions from this chapter

The Fibonacci sequence \(\\{1,1,2,3,5,8,13, \ldots\\}\) is generated by the recurrence relation \(f_{n+1}=f_{n}+f_{n-1},\) for \(n=1,2,3, \ldots,\) where \(f_{0}=1, f_{1}=1\). a. It can be shown that the sequence of ratios of successive terms of the sequence \(\left\\{\frac{f_{n+1}}{f_{n}}\right\\}\) has a limit \(\varphi .\) Divide both sides of the recurrence relation by \(f_{n},\) take the limit as \(n \rightarrow \infty,\) and show that \(\varphi=\lim _{n \rightarrow \infty} \frac{f_{n+1}}{f_{n}}=\frac{1+\sqrt{5}}{2} \approx 1.618\). b. Show that \(\lim _{n \rightarrow \infty} \frac{f_{n-1}}{f_{n+1}}=1-\frac{1}{\varphi} \approx 0.382\). c. Now consider the harmonic series and group terms as follows: $$\sum_{k=1}^{\infty} \frac{1}{k}=1+\frac{1}{2}+\frac{1}{3}+\left(\frac{1}{4}+\frac{1}{5}\right)+\left(\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)$$ $$+\left(\frac{1}{9}+\cdots+\frac{1}{13}\right)+\cdots$$ With the Fibonacci sequence in mind, show that $$\sum_{k=1}^{\infty} \frac{1}{k} \geq 1+\frac{1}{2}+\frac{1}{3}+\frac{2}{5}+\frac{3}{8}+\frac{5}{13}+\cdots=1+\sum_{k=1}^{\infty} \frac{f_{k-1}}{f_{k+1}}.$$ d. Use part (b) to conclude that the harmonic series diverges. (Source: The College Mathematics Journal, 43, May 2012)

Use Theorem 8.6 to find the limit of the following sequences or state that they diverge. $$\left\\{\frac{n^{10}}{\ln ^{1000} n}\right\\}$$

In \(1978,\) in an effort to reduce population growth, China instituted a policy that allows only one child per family. One unintended consequence has been that, because of a cultural bias toward sons, China now has many more young boys than girls. To solve this problem, some people have suggested replacing the one-child policy with a one-son policy: A family may have children until a boy is born. Suppose that the one-son policy were implemented and that natural birth rates remained the same (half boys and half girls). Using geometric series, compare the total number of children under the two policies.

In Section \(8.3,\) we established that the geometric series \(\sum r^{k}\) converges provided \(|r| < 1\). Notice that if \(-1 < r<0,\) the geometric series is also an alternating series. Use the Alternating Series Test to show that for \(-1 < r <0\), the series \(\sum r^{k}\) converges.

Consider the following sequences defined by a recurrence relation. Use a calculator, analytical methods, and/or graphing to make a conjecture about the limit of the sequence or state that the sequence diverges. $$a_{n+1}=\frac{1}{2}\left(a_{n}+2 / a_{n}\right) ; a_{0}=2$$

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