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Chapter 9: Problem 44

Find the limit of the following sequences or determine that the limit does not exist. Verify your result with a graphing utility. $$a_{n}=\cot \left(\frac{n \pi}{2 n+2}\right)$$

Short Answer

Expert verified
Answer: The limit of the sequence does not exist as \(n \to \infty\).

Step by step solution

01

Rewrite the cotangent function as a reciprocal of the tangent function

We have the sequence $$a_{n}=\cot \left(\frac{n \pi}{2 n+2}\right) \implies a_n = \frac{1}{\tan\left(\frac{n \pi}{2 n+2}\right)}.$$
02

Evaluate the input of the tangent function when n approaches infinity

As \(n \to \infty\), the input of the tangent function \(\frac{n \pi}{2 n+2}\) also approaches a limiting value. Let's evaluate this limit: $$\lim_{n \to \infty} \frac{n \pi}{2 n+2}=\lim_{n \to \infty} \frac{\pi}{1+\frac{1}{n}}.$$ Since \(\lim_{n \to \infty} \frac{1}{n} = 0\), we have $$\lim_{n \to \infty} \frac{\pi}{1+\frac{1}{n}} = \frac{\pi}{1+0} = \pi.$$
03

Evaluate the limit of the sequence

Now we can evaluate the limit of our sequence as \(n \to \infty\): $$\lim_{n \to \infty} a_{n}=\lim_{n \to \infty}\frac{1}{\tan\left(\frac{n \pi}{2 n+2}\right)} = \frac{1}{\tan\left(\lim_{n \to \infty} \frac{n \pi}{2 n+2}\right)} = \frac{1}{\tan(\pi)}.$$ Since \(\tan(\pi) = 0\), we have \(\frac{1}{\tan(\pi)} = \frac{1}{0}\). The limit does not exist in this case, as the sequence \(a_{n}\) is not converging to any finite value.
04

Verify the result with a graphing utility

Use a graphing utility to plot the sequence \(a_{n}=\cot\left(\frac{n \pi}{2 n+2}\right)\) for various values of \(n\). Observe that the values of the sequence oscillate between positive and negative infinity (i.e. increasing the values of \(n\) makes it approach either a vertical asymptote or zero). This confirms that the limit does not exist as \(n \to \infty\).

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Most popular questions from this chapter

Determine whether the following series converge absolutely or conditionally, or diverge. $$\sum_{k=1}^{\infty} \frac{(-1)^{k+1} e^{k}}{(k+1) !}$$

Evaluate the limit of the following sequences. $$a_{n}=\frac{6^{n}+3^{n}}{6^{n}+n^{100}}$$

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

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

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.
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