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Chapter 3: Problem 8

Determine the limits of the following. (a) \(\left((3 n)^{1 / 2 n}\right)\), (b) \(\left((1+1 / 2 n)^{3 n}\right)\).

Short Answer

Expert verified
The limit of \((3n)^{1/2n}\) as n approaches infinity is 1 and the limit of \((1 + 1/(2n))^{3n}\) as n approaches infinity is \(e^{-3/2}\).

Step by step solution

01

Analyze the first function

Consider the function \((3n)^{1 / 2n}\). As n approaches infinity, the base \(3n\) grows infinitely large and the exponent \(1 / 2n\) tends to zero, because it is a fraction where the denominator becomes large. This results in an indeterminate form. To solve this we can take the natural logarithm on both sides and then apply limit properties.
02

Solve for the limit of first function

Taking the natural logarithm, obtain \(\ln\left((3n)^{1/2n}\right) = \frac{1}{2n}\ln(3n)\). Now, if we take the limit as n approaches infinity, we get the indeterminate form \(\frac{0}{0}\), to find the limit in this situation, L'Hopital's Rule should be applied, which states that this limit is equal to the limit of the ratio between the derivatives of the top and the bottom functions. So, applying L'Hopital's Rule we have \(\lim_{n \to \infty}(\frac{1}{2n}\ln(3n)) = \lim_{n \to \infty}(\frac{d(\ln(3n))}{d(1/2n)}) = \lim_{n \to \infty}(\frac{3/n}{-1/(4n^{2})})\), simplifying we have \(\lim_{n \to \infty}(-\frac{12}{n}) = 0\). Tus, \( \lim_{n \to \infty}((3n)^{1/2n}) = e^0 = 1\).
03

Analyze the second function

Now, consider the function \((1 + 1/(2n))^{3n}\). As before, as n approaches infinity, the base \((1+1/(2n))\) tends to 1 and the exponent \(3n\) tends to infinity. This results once again in an indeterminate form that will require similar steps as in the first function to solve it.
04

Solve for the limit of the second function

Let's take the natural logarithm of the function, which gives us \(\ln \left((1+1/(2n))^{3n}\right) = 3n \ln(1 + 1/(2n))\). When we apply the limit as n approaches infinity here, we obtain once again an indeterminate form of \(0*鈭瀄). To apply L'Hopital's Rule, we express the form \(0*鈭瀄) as \(\frac{鈭瀩{鈭瀩\), that way we obtain \(\lim_{n\to\infty} \frac{n}{1/(3 \ln(1+1/(2n))))}\). Now, applying L'Hopital's Rule we get \(\lim_{n \to \infty}(\frac{d(n)}{d(1/(3 \ln(1+1/(2n))))}) = \lim_{n \to \infty}(\frac{1}{-3/(2n^{2}(1 + 1/(2n)))}) = \lim_{n \to \infty}(-3/2 * n^{2}*(1 + 1/(2n))) = -3/2\). Thus, \( \lim_{n \to \infty}(1 + 1/(2n))^{3n} = e^{-3/2}\).

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Most popular questions from this chapter

Let \(\sum a_{n}\) be a given series and let \(\sum b_{n}\) be the series in which the terms are the same and in the same order as in \(\sum a_{n}\) except that the terms for which \(a_{n}=0\) have been omitted. Show that \(\sum a_{n}\) converges to \(A\) if and only if \(\sum b_{n}\) converges to \(A\).

Let \(\left(I_{n}\right)\) be a nested sequence of closed bounded intervals. For each \(n \in \mathbb{N}\), let \(x_{n} \in I_{n} .\) Use the Bolzano-Weierstrass Theorem to give a proof of the Nested Intervals Property 2.5.2.

Let \(\left(x_{n}\right)\) be a Cauchy sequence such that \(x_{n}\) is an integer for every \(n \in \mathbb{N}\). Show that \(\left(x_{n}\right)\) is. ultimately constant.

Show that the following sequences are not convergent. (a) \(\left(2^{n}\right)\), (b) \(\left((-1)^{n} n^{2}\right)\).

Establish the convergence and find the limits of the following sequences: (a) \(\left(\left(1+1 / n^{2}\right)^{n^{2}}\right)\). (b) \(\left((1+1 / 2 n)^{n}\right)\), (c) \(\left(\left(1+1 / n^{2}\right)^{2 n^{2}}\right)\). (d) \(\left((1+2 / n)^{n}\right)\).
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