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Evaluate the limit of the following sequences or state that the limit does not exist. $$a_{n}=\frac{n^{8}+n^{7}}{n^{7}+n^{8} \ln n}$$

Short Answer

Expert verified
Answer: The limit of the sequence as n approaches infinity is 0.

Step by step solution

01

Analyze given sequence

Our goal here is to analyze the given sequence and understand its behavior as n approaches infinity. The sequence is: $$a_{n}=\frac{n^{8}+n^{7}}{n^{7}+n^{8} \ln n}$$
02

Simplify expression

We can simplify this expression by factoring out the common term \(n^7\) from both numerator and denominator, which will make it easier to identify the limit as n approaches infinity: $$a_{n}=\frac{n^7(n + 1)}{n^7(1 + n \ln n)} = \frac{n + 1}{1 + n \ln n}$$ Now the expression is simplified, and we can proceed to evaluate the limit.
03

Check limit behavior

In order to check the limit behavior, we will examine the behavior of the simplified expression as n approaches infinity: $$\lim_{n \to \infty} \frac{n+1}{1+n \ln n}$$ Now we can plainly see that both the numerator and the denominator tend to infinity as n grows.
04

Apply L'Hôpital's Rule

Since both the numerator and the denominator tend to infinity, we have an indeterminate form, and thus we can apply L'Hôpital's Rule. We compute the derivatives of both the numerator and the denominator: $$\frac{d}{dn}(n + 1) = 1$$ $$\frac{d}{dn}(1 + n \ln n) = 1 + \ln n$$ Now we express the limit in terms of the derivatives: $$\lim_{n \to \infty} \frac{1}{1 + \ln n}$$
05

Evaluate final limit

Finally, we evaluate this remaining limit as n approaches infinity: $$\lim_{n \to \infty} \frac{1}{1 + \ln n}$$ As n goes to infinity, the expression \(\ln n\) will also go to infinity. Therefore, the denominator will go to infinity, and the whole fraction will go to 0. Thus, the final answer is: $$\lim_{n \to \infty} a_n = 0$$

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Key Concepts

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

L'Hôpital's Rule
L'Hôpital's Rule is a mathematical tool used to solve limits that present indeterminate forms such as \( \frac{\infty}{\infty} \) or \( \frac{0}{0} \). It is particularly useful when both the numerator and the denominator of a fraction approach infinity or zero as the variable approaches a certain value. This rule allows you to differentiate the top and bottom separately until a determinate form is achieved.
  • To apply L'Hôpital's Rule, ensure the limit you are evaluating results in an indeterminate form.
  • Differentiation is essential. Compute the derivative of both the numerator and the denominator.
  • Re-evaluate the limit with these new expressions.
L'Hôpital's Rule simplifies complex rational expressions and helps find precise limit values. In our exercise, we applied this rule after simplifying the sequence and identifying that both the numerator (\(n+1\)) and the denominator (\(1+n \ln n\)) tended towards infinity. After applying the rule, the limit became a simple rational function which was easy to evaluate.
Limit Evaluation
Limit Evaluation is the process of finding the value a sequence approaches as the variable moves toward a particular point, which is frequently infinity. Assessing limits can help understand the behavior of expressions, especially when dealing with sequences and functions that describe real-world situations.
In our problem, we needed to evaluate the limit of the sequence \(a_n = \frac{n^8 + n^7}{n^7 + n^8 \ln n}\) as \(n\) approaches infinity. This involved several steps:
  • Simplifying the expression by factoring out common terms to make the limit more manageable.
  • Examining the result to determine the form of the expression (e.g., if it leads to \(\infty - \infty\) or \(\frac{0}{0}\)).
After simplification, we were left with \(\frac{n+1}{1+n \ln n}\), which clearly indicated an \(\frac{\infty}{\infty}\) form. Thus, this example exemplified the importance of correct simplification in limit evaluation, setting up the need for L'Hôpital's Rule.
Indeterminate Forms
Indeterminate Forms are results of limits where standard algebraic methods don't apply directly. Common indeterminate forms include \(\frac{0}{0}\), \(\frac{\infty}{\infty}\), and other expressions like \(0 \cdot \infty\) and \(\infty - \infty\). They suggest that a straightforward evaluation is not possible, as the result could be anything without additional analysis.

Why Indeterminate Forms Matter

An indeterminate form means your sequence or function isn’t clearly trending towards a limit as expected. Both our numerator and denominator were growing towards infinity in the example problem, which resulted in \(\frac{\infty}{\infty}\), a classic indeterminate form. This signals more advanced methods, such as L'Hôpital's Rule, are required to evaluate the limit.
  • The appearance of an indeterminate form confirms the need for further investigation or rule application.
  • Understanding them is crucial to correctly applying calculus to real-world problems.
Indeterminate forms must be recognized swiftly to apply the correct mathematical strategies efficiently, as was done in this problem to isolate and solve the limit successfully.

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Most popular questions from this chapter

Imagine that the government of a small community decides to give a total of \(\$ W\), distributed equally, to all its citizens. Suppose that each month each citizen saves a fraction \(p\) of his or her new wealth and spends the remaining \(1-p\) in the community. Assume no money leaves or enters the community, and all the spent money is redistributed throughout the community. a. If this cycle of saving and spending continues for many months, how much money is ultimately spent? Specifically, by what factor is the initial investment of \(\$ W\) increased (in terms of \(p\) )? Economists refer to this increase in the investment as the multiplier effect. b. Evaluate the limits \(p \rightarrow 0\) and \(p \rightarrow 1,\) and interpret their meanings.

Use the ideas of Exercise 88 to evaluate the following infinite products. $$\text { a. } \prod_{k=0}^{\infty} e^{1 / 2^{k}}=e \cdot e^{1 / 2} \cdot e^{1 / 4} \cdot e^{1 / 8} \dots$$ $$\text { b. } \prod_{k=2}^{\infty}\left(1-\frac{1}{k}\right)=\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdot \frac{4}{5} \cdots$$

Prove that if \(\sum a_{k}\) diverges, then \(\sum c a_{k}\) also diverges, where \(c \neq 0\) is a constant.

In the following exercises, two sequences are given, one of which initially has smaller values, but eventually "overtakes" the other sequence. Find the sequence with the larger growth rate and the value of \(n\) at which it overtakes the other sequence. $$a_{n}=n ! \text { and } b_{n}=n^{0.7 n}, n \geq 2$$

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)

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