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Chapter 3: Problem 15

Show that \(\lim \left((2 n)^{1 / n}\right)=1\).

Short Answer

Expert verified
The limit of the sequence \( (2n)^{1/n} \) as \( n \) approaches infinity is 1.

Step by step solution

01

Identify the Sequence

The given sequence is \( (2n)^{1/n} \). You need to find the limit as \( n \) approaches infinity.
02

Rewrite the Sequence

You can rewrite this sequence as \( e^{ \left( \frac{1}{n} \right) \ln(2n) } \) to make it easier to work with.
03

Simplify the Exponent

Let's simplify the exponent. As \( n \) approaches infinity, the term \( \frac{1}{n} \) approaches 0. Therefore, the limit of the sequence simplifies to \( e^{0} \) as \( n \) approaches infinity.
04

Calculate the Limit

Because \( e^{0} \) is equal to 1, the limit of the sequence \( (2n)^{1/n} \) as \( n \) approaches infinity is 1.

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Key Concepts

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

Sequence
A sequence is like a list of numbers written in a particular order. Each number in this order is called a 'term'. When dealing with sequences, you often look at how these terms behave as you keep going further in the list. In this exercise, the sequence given is \((2n)^{1/n}\). This means each term is taken by raising \(2n\) to the power of \(1/n\).

Sequences are important in calculus and mathematics because they help us understand trends and patterns as you move towards infinity. Here, the goal is to find the limit of the sequence as \(n\) becomes very large, or approaches infinity. It's like seeing what the list of numbers is pointing towards as you go further along.
Exponential Function
The exponential function is a mathematical function denoted as \(e^x\), where \(e\) is approximately 2.71828. It is a unique and important function because it grows very quickly. In this solution, we rewrite the sequence \((2n)^{1/n}\) using the property of logarithms: \(a^{b} = e^{b \ln(a)}\).

This means \((2n)^{1/n}\) becomes \(e^{(1/n) \ln(2n)}\). Rewriting it this way makes it easier to handle especially when calculating limits because exponents and \(e\) have helpful rules and behaviors.
  • The base \(e\) makes calculating derivatives and integrals more straightforward.
  • Using \(e^{x}\) allows us to simplify complex exponent expressions.
Simplifying the exponent, \(\frac{1}{n} \ln(2n)\), helps us see how the sequence behaves as \(n\) approaches infinity.
Infinite Limits
Infinite limits deal with understanding what happens when the values in a sequence or function become very large. In this context, we look at \(n\) approaching infinity. For the sequence \((2n)^{1/n}\), determining the limit as \(n\) approaches infinity involves studying the behavior of expressions like \(\frac{1}{n} \ln(2n)\).

As \(n\) gets larger, \(\frac{1}{n}\) becomes smaller, and this significantly impacts the result. When \(\frac{1}{n}\) approaches zero, the exponent \(e^{\left(\frac{1}{n}\right) \ln(2n)}\) approaches \(e^0\).
  • Understanding infinite limits is crucial when you're trying to determine what value a sequence 'aims' for.
  • Calculating limits helps in graphing functions, studying series, and solving advanced calculus problems.
In our exercise, this analysis showed that the limit of the sequence is indeed 1.

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Most popular questions from this chapter

If \(\sum a_{n}\) with \(a_{n}>0\) is convergent, then is \(\sum \sqrt{a_{n}}\) always convergent? Either prove it or give a counterexample.

If \(y_{1}2\), show that \(\left(y_{n}\right)\) is convergent. What is its limit?

Establish the convergence or the divergence of the sequence \(\left(y_{n}\right)\), where $$y_{n}:=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2 n} \quad \text { for } \quad n \in \mathbb{N} .$$

Let \(x_{1} \geq 2\) and \(x_{n+1}:=1+\sqrt{x_{n}-1}\) for \(n \in \mathbb{N}\). Show that \(\left(x_{n}\right)\) is decreasing and bounded below by \(2 .\) Find the limit.

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