Question

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

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

Step-by-step solution

Rewrite the cotangent function as a reciprocal of the tangent function

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

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 $$\lim_{n \to \infty} \frac{\pi}{1+\frac{1}{n}} = \frac{\pi}{1+0} = \pi.$$

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.

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 \(n \to \infty\).