/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 25 Find the limit of the following ... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the limit of the following sequences or determine that the limit does not exist. $$\left\\{\left(\frac{1}{n}\right)^{1 / n}\right\\}$$

Short Answer

Expert verified
Answer: The limit of the sequence is 1.

Step by step solution

01

Identify the sequence formula

The given sequence is: $$a_n = \left(\frac{1}{n}\right)^{1/n}$$
02

Use logarithms

Since the expression is an exponential function of n, we can use logarithms to simplify it. We will apply the natural logarithm to both sides: $$\ln a_n = \ln\left(\frac{1}{n}\right)^{1/n}$$ Using properties of logarithms, we can rewrite the expression as: $$\ln a_n = \frac{1}{n} \ln\left(\frac{1}{n}\right)$$
03

Find the limit of log sequence

Before we find the limit of the original sequence, let's find the limit of the transformed sequence \(\ln a_n\): $$\lim_{n \to \infty} \ln a_n = \lim_{n \to \infty} \frac{1}{n} \ln\left(\frac{1}{n}\right)$$
04

Substitute using L'Hôpital's rule

The transformed limit has the indeterminate form \(\frac{0}{\infty}\). Thus, we can apply L'Hôpital's rule: $$\lim_{n \to \infty} \ln a_n = \lim_{n \to \infty} \frac{\frac{d}{dn} \ln\left(\frac{1}{n}\right)}{\frac{d}{dn} n}$$ Differentiating the numerator and denominator respectively: $$\lim_{n \to \infty} \ln a_n = \lim_{n \to \infty} \frac{-\frac{1}{n^2}}{1}$$
05

Evaluate the limit of log sequence

We can now evaluate this limit as n approaches infinity: $$\lim_{n \to \infty} \ln a_n = \lim_{n \to \infty} -\frac{1}{n^2} = 0$$
06

Use the continuity of exponential functions

Now, we can use the continuity property of exponential functions to find the limit of the original sequence: $$\lim_{n \to \infty} a_n = \lim_{n \to \infty} e^{\ln a_n} = e^{\lim_{n \to \infty} \ln a_n}$$ Substitute the limit we calculated in Step 5: $$\lim_{n \to \infty} a_n = e^0$$
07

Get the limit result

Now, evaluate the expression for the limit: $$\lim_{n \to \infty} a_n = e^0 = 1$$ So, the limit of the given sequence exists and is equal to 1.

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Key Concepts

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

L'Hôpital's Rule
Understanding L'Hôpital's rule is critical when dealing with limits that result in indeterminate forms like \(0/0\) or \(\infty/\infty\). The rule is a way to evaluate such limits with greater ease, providing a method to find the limit of ratios of functions. According to this rule, if we have:\[\lim_{x \to c} \frac{f(x)}{g(x)}\] and both \(\frac{d}{dx}f(x)\) and \(\frac{d}{dx}g(x)\) are continuous near \(c\), and if the limit results in an indeterminate form, we can differentiate both the numerator and denominator separately and then take the limit of the ratio of these derivatives. It is important to note that this process might need to be repeated if the resulting limit still leads to an indeterminate form. This rule is particularly useful in your study of calculus and in exercises such as the given sequence problem.

When applying L'Hôpital's rule to the sequence \(\ln a_n\), we differentiate both the numerator \(\ln(1/n)\) and denominator \((n)\) separately and take the limit. This simplifies the evaluation, giving us a clear path to find that the limit of the logarithmic sequence, and hence the original sequence, is 1.
Properties of Logarithms
Logarithms have certain algebraic properties that make them invaluable tools in solving limit problems, especially when dealing with sequences that are expressed in terms of exponents. Some fundamental properties include:
  • \(\log(a \cdot b) = \log(a) + \log(b)\)
  • \(\log(a / b) = \log(a) - \log(b)\)
  • \(\log(a^k) = k \cdot \log(a)\)
  • \(\log_b(a) = \frac{\log_c(a)}{\log_c(b)}\) (change of base formula)
Applying these properties allows for the simplification of the sequence \(a_n \), expressed as the nth root of 1/n, into a more manageable form. Specifically, when the natural logarithm is applied, the power rule for logarithms is used, bringing the exponent in front as a multiplier. This transformation simplifies the original problem, allowing us to apply other mathematical tools such as L'Hôpital's rule. When you encounter sequences or problems that are not easily solvable in their current form, consider whether such properties of logarithms can be applied to aid in finding the solution.

It's essential for students to practice using these properties not only to simplify complex expressions but also to develop a deeper understanding of logarithmic functions and their applications in different mathematical contexts.
Continuity of Exponential Functions
Exponential functions are widely used in various fields due to their continuous nature over their entire domain, which in most cases is all real numbers. This makes them highly predictable and smooth, which is important for solving limits involving exponents. An exponential function has the general form \(f(x) = a^x\), where \(a > 0\) is a constant. For the natural exponential function \(e^x\), continuity means that the function has no breaks, jumps, or holes in its graph. This property of exponential functions enables us to apply the limit of the logarithm of a sequence directly to the original sequence itself.

When we reached the point in the step-by-step solution where the limit of the logarithmic sequence \(\ln a_n\) is 0, the continuity of the function \(e^x\) guaranteed that the limit could be accurately evaluated using the exponential function. Substituting the limit back into the exponential form results in \(e^0\), which simplifies to 1. Therefore, our original sequence's limit exists and is 1. The continuity of exponential functions thus ensures a seamless transition from logarithmic form back to the original sequence, making the overall evaluation process much smoother.

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Most popular questions from this chapter

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 \(10^{-3}\).

Prove that if \(\sum a_{k}\) diverges, then \(\sum c a_{k}\) also diverges, where \(c \neq 0\) is a constant.

Evaluate the limit of the following sequences or state that the limit does not exist. $$a_{n}=\frac{7^{n}}{n^{7} 5^{n}}$$

Stirling's formula Complete the following steps to find the values of \(p>0\) for which the series \(\sum_{k=1}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdots(2 k-1)}{p^{k} k !}\) converges. a. Use the Ratio Test to show that \(\sum_{k=1}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdots(2 k-1)}{p^{k} k !}\) converges for \(p>2\). b. Use Stirling's formula, \(k !=\sqrt{2 \pi k} k^{k} e^{-k}\) for large \(k,\) to determine whether the series converges when \(p=2\). (Hint: \(1 \cdot 3 \cdot 5 \cdots(2 k-1)=\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdots(2 k-1) 2 k}{2 \cdot 4 \cdot 6 \cdots 2 k}\) (See the Guided Project Stirling's formula and \(n\) ? for more on this topic.)

Consider the geometric series \(S=\sum_{k=0}^{\infty} r^{k}\) which has the value \(1 /(1-r)\) provided \(|r|<1\). Let \(S_{n}=\sum_{k=0}^{n-1} r^{k}=\frac{1-r^{n}}{1-r}\) be the sum of the first \(n\) terms. The magnitude of the remainder \(R_{n}\) is the error in approximating \(S\) by \(S_{n} .\) Show that $$ R_{n}=S-S_{n}=\frac{r^{n}}{1-r} $$

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