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Chapter 8: Problem 83

Evaluate the limit of the following sequences or state that the limit does not exist. $$a_{n}=\frac{6^{n}+3^{n}}{6^{n}+n^{100}}$$

Short Answer

Expert verified
Answer: The limit of the sequence \(a_{n}\) as n approaches infinity is 1.

Step by step solution

01

Identify the dominant terms

The dominant terms are the terms with the highest exponent of n or the terms with the most significant contribution as n approaches infinity. In this case, the dominant term in the numerator is \(6^{n}\) since it grows much faster than \(3^{n}\). Similarly, the dominant term in the denominator is \(6^{n}\) as \(n^{100}\) grows more slowly in comparison. #Step 2: Divide the numerator and the denominator by the dominant term#
02

Divide by the dominant term

Divide both the numerator and the denominator by \(6^{n}\). $$ \frac{ \frac{6^{n} + 3^{n}}{6^{n}} }{ \frac{6^{n} + n^{100}}{6^{n}} } $$ #Step 3: Simplify the expression#
03

Simplify the expression

After dividing by \(6^{n}\), the expression becomes: $$ \frac{ 1 + \left(\frac{3}{6}\right)^{n} }{ 1 + \frac{n^{100}}{6^{n}} } $$ #Step 4: Find the limit#
04

Find the limit as n approaches infinity

As n approaches infinity, \(\left(\frac{3}{6}\right)^{n}\) approaches 0 because \((\frac{1}{2})^{n}\) goes to 0 while \(n\to\infty\). Similarly, the term \(\frac{n^{100}}{6^{n}}\) approaches 0 since the exponential function grows much faster than the polynomial function. The limit becomes: $$ \lim_{n \to \infty} \frac{ 1 + \left(\frac{3}{6}\right)^{n} }{ 1 + \frac{n^{100}}{6^{n}} } = \frac{ 1 + 0 }{ 1 + 0 } = \frac{1}{1} = 1 $$ The limit of the sequence \(a_{n}\) as n approaches infinity is 1.

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Key Concepts

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

Limits
Calculus often involves finding the limit of a sequence or function as it approaches a particular point or infinity. A limit examines the behavior of the terms in a sequence or the values a function produces when the input gets very large or very close to a specific point.

In this exercise, we are dealing with the sequence \(a_n = \frac{6^n + 3^n}{6^n + n^{100}}\). We are asked to find what happens to this sequence as \(n\) becomes very large.
  • A crucial first step is determining which terms grow fastest as \(n\) increases because these terms influence the limit the most.
  • "Dominant terms" play a key role, as they dictate the sequence's behavior at infinity.
By identifying and simplifying with these dominant terms, we can find that the limiting value of the sequence is 1 as \(n\) goes to infinity. Limits help us understand the long-term behavior of sequences and functions, giving insights into their stability and end behaviors.
Sequences
Sequences are ordered lists of numbers defined by an explicit formula. Each number in the sequence is called a term. The sequence can be finite or infinite depending on how we define it.

For example, the sequence given in this problem, \(a_n = \frac{6^n + 3^n}{6^n + n^{100}}\), is an infinite sequence as it progresses to infinity.
  • Each term in the sequence is identified with an integer index \(n\).
  • The behavior of such a sequence as \(n\) becomes extremely large is often of interest, as it can show trends and asymptotic behavior.
Analyzing sequences extensively is crucial in calculus and mathematical analysis. This is because they offer a foundation for understanding series, calculus limits, and real-valued functions. In this exercise, observing how sequence terms simplify as \(n\) gets larger provides a glimpse into limit concepts and asymptotic properties.
Asymptotic Analysis
Asymptotic analysis is a method used in calculus and computer science to describe the behavior of functions or sequences as the input or index approaches infinity.

For the given exercise, asymptotic analysis involves understanding how the dominant term \(6^n\) dominates the sequence's growth, allowing us to simplify terms like \(\left(\frac{3}{6}\right)^n\) and \(\frac{n^{100}}{6^n}\) to zero.
  • This technique relies heavily on dominance; exponential terms typically overshadow polynomial terms as \(n\) becomes very large.
  • By focusing on the dominant behaviors, it becomes easier to predict and understand the sequence's behavior without exact calculations for each term.
Asymptotic analysis is a powerful technique for finding limits, understanding complexity in algorithms, and simplifying complex expressions. In this scenario, the attention is on catching how each term's magnitude influences the entire expression's behavior, helping us compute limits seamlessly as \(n\) approaches infinity.

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Most popular questions from this chapter

An insulated window consists of two parallel panes of glass with a small spacing between them. Suppose that each pane reflects a fraction \(p\) of the incoming light and transmits the remaining light. Considering all reflections of light between the panes, what fraction of the incoming light is ultimately transmitted by the window? Assume the amount of incoming light is 1.

Assume that \(\sum_{k=1}^{\infty} \frac{1}{k^{2}}=\frac{\pi^{2}}{6}\) (Exercises 65 and 66 ) and that the terms of this series may be rearranged without changing the value of the series. Determine the sum of the reciprocals of the squares of the odd positive integers.

Give an argument similar to that given in the text for the harmonic series to show that \(\sum_{k=1}^{\infty} \frac{1}{\sqrt{k}}\) diverges.

The Riemann zeta function is the subject of extensive research and is associated with several renowned unsolved problems. It is defined by \(\zeta(x)=\sum_{k=1}^{\infty} \frac{1}{k^{x}} .\) When \(x\) is a real number, the zeta function becomes a \(p\) -series. For even positive integers \(p,\) the value of \(\zeta(p)\) is known exactly. For example, $$\sum_{k=1}^{\infty} \frac{1}{k^{2}}=\frac{\pi^{2}}{6}, \quad \sum_{k=1}^{\infty} \frac{1}{k^{4}}=\frac{\pi^{4}}{90}, \quad \text { and } \quad \sum_{k=1}^{\infty} \frac{1}{k^{6}}=\frac{\pi^{6}}{945}, \ldots$$ Use the estimation techniques described in the text to approximate \(\zeta(3)\) and \(\zeta(5)\) (whose values are not known exactly) with a remainder less than \(10^{-3}\).

Here is a fascinating (unsolved) problem known as the hailstone problem (or the Ulam Conjecture or the Collatz Conjecture). It involves sequences in two different ways. First, choose a positive integer \(N\) and call it \(a_{0} .\) This is the seed of a sequence. The rest of the sequence is generated as follows: For \(n=0,1,2, \ldots\) $$a_{n+1}=\left\\{\begin{array}{ll} a_{n} / 2 & \text { if } a_{n} \text { is even } \\ 3 a_{n}+1 & \text { if } a_{n} \text { is odd .} \end{array}\right.$$ However, if \(a_{n}=1\) for any \(n,\) then the sequence terminates. a. Compute the sequence that results from the seeds \(N=2,3\), \(4, \ldots, 10 .\) You should verify that in all these cases, the sequence eventually terminates. The hailstone conjecture (still unproved) states that for all positive integers \(N\), the sequence terminates after a finite number of terms. b. Now define the hailstone sequence \(\left\\{H_{k}\right\\},\) which is the number of terms needed for the sequence \(\left\\{a_{n}\right\\}\) to terminate starting with a seed of \(k\). Verify that \(H_{2}=1, H_{3}=7\), and \(H_{4}=2\). c. Plot as many terms of the hailstone sequence as is feasible. How did the sequence get its name? Does the conjecture appear to be true?
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