Question

Evaluate the following limits: (i) \(\lim _{n \rightarrow x}\left[\tan \frac{\pi}{2 n} \tan \frac{2 \pi}{2 n} \tan \frac{3 \pi}{2 n} \ldots . \tan \frac{n \pi}{2 n}\right]^{1 / n}\) (ii) \(\lim _{n \rightarrow x}\left[\left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right)^{1 / 2}\left(1+\frac{3}{n}\right)^{1 / 3} \ldots\left(1+\frac{n}{n}\right)^{v_{n}}\right]\)

Short answer

Based on the step-by-step solution provided: Question: Determine the limits of the following expressions as n approaches infinity: (i) \(\lim_{n\rightarrow\infty}\left[\tan\frac{\pi}{2n}\tan\frac{2\pi}{2n}\ldots\tan\frac{n\pi}{2n}\right]^{1/n}\) (ii) \(\lim_{n\rightarrow\infty}\left[\prod_{k=1}^{n}\left(1+\frac{k}{n}\right)^{1/k}\right]\) Answer: (i) The limit of the first expression as n approaches infinity is 1. (ii) The limit of the second expression as n approaches infinity does not exist since the expression grows without bounds.

Step-by-step solution

Evaluate the limit for the first function (i)

First, note that the expression inside the limit can be rewritten as the product of tangent functions. The limit can be expressed as: \(\lim_{n\rightarrow\infty}\left[\tan\frac{\pi}{2n}\tan\frac{2\pi}{2n}\ldots\tan\frac{n\pi}{2n}\right]^{1/n}\) Notice that the angles of the tangent functions are in arithmetic progression with a common difference of \(\frac{\pi}{2n}\).

Use the tangent multiplication formula

We use the formula \(\tan a \tan b = \frac{\sin (a + b)}{\cos a \cos b}\). Applying this recursively to the product of all the terms, we get: \(\lim_{n\rightarrow\infty}\left[\frac{\sin\frac{3\pi}{2n}}{\cos\frac{\pi}{2n}\cos\frac{2\pi}{2n}}\times\frac{\sin\frac{5\pi}{2n}}{\cos\frac{3\pi}{2n}\cos\frac{4\pi}{2n}}\times\ldots\times\frac{\sin\frac{(2n-1)\pi}{2n}}{\cos\frac{(n-1)\pi}{2n}\cos\frac{n\pi}{2n}}\right]^{1/n}\)

Combine terms and simplify

Notice that it can be simplified as: \(\lim_{n\rightarrow\infty}\left[\frac{\sin\frac{3\pi}{2n}}{\cos\frac{\pi}{2n}}\times\frac{\sin\frac{5\pi}{2n}}{\cos\frac{2\pi}{2n}}\times\ldots\times\frac{\sin\frac{(2n-1)\pi}{2n}}{\cos\frac{n\pi}{2n}}\right]^{1/n}\) Now, we can easily find that the limit is \(1\).

Evaluate the limit for the second function (ii)

We rewrite the limit in a more general form: \(\lim_{n\rightarrow\infty}\left[\prod_{k=1}^{n}\left(1+\frac{k}{n}\right)^{1/k}\right]\) Notice that this expression can be rewritten as a product of binomial expressions raised to a fraction power.

Take the natural logarithm of the expression

To find the limit, it's helpful to take the natural logarithm of the expression: \(\lim_{n\rightarrow\infty}\ln\left[\prod_{k=1}^{n}\left(1+\frac{k}{n}\right)^{1/k}\right]\) Using the properties of logarithms, this becomes: \(\lim_{n\rightarrow\infty}\sum_{k=1}^{n}\frac{1}{k}\ln\left(1+\frac{k}{n}\right)\)

Convert the sum into an integral

As n approaches infinity, the sum can be approximated by an integral: \(\lim_{n\rightarrow\infty}\frac{1}{n}\int_{1}^{n}\frac{\ln(1+x/n)}{x}dx\)

Evaluate the integral and find the limit

The integral evaluates to \([\ln(1+\frac{x}{n})]_{1}^{n}\), which simplifies to: \(\ln(1+(n-1))-\ln2 = \ln(n)-\ln2\) Hence, the limit of the original expression is e raised to the power of the limit of the natural logarithm expression, which is: \(\lim_{n\rightarrow\infty}e^{\ln(n)-\ln2}\) This limit does not exist since the expression grows without bounds.