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Chapter 11: Problem 40

Find the limit of the sequence. $$\lim _{n \rightarrow \infty} n^{\sin (\pi / n)}$$

Short Answer

Expert verified
The short answer is: \(\lim _{n \rightarrow \infty} n^{\sin (\pi / n)} = \frac{1}{e}\).

Step by step solution

01

Rewrite expression with natural logarithm

To tackle this limit, rewrite the given expression in terms of natural logarithms as follows: \[\lim _{n \rightarrow \infty} \ln\left(n^{\sin (\pi / n)}\right)\]
02

Apply logarithm properties

Use the properties of logarithms to simplify the expression: \[\lim _{n \rightarrow \infty} \left[\sin (\pi / n) \cdot \ln (n)\right]\]
03

Apply L'Hopital's Rule

Now, consider the limit as a fraction so we can use L'Hopital's Rule. The expression can be re-written as: \[\lim _{n \rightarrow \infty} \frac{\ln (n)}{\frac{1}{\sin (\pi / n)}}\] Let \(x = \frac{\pi}{n}\), then as \(n \rightarrow \infty\), \(x \rightarrow 0\). The limit becomes: \[\lim _{x \rightarrow 0} \frac{\ln (\frac{\pi}{x})}{\frac{1}{\sin (x)}}\] Next, recognize that this limit is of the indeterminate form \(\frac{0}{0}\) and apply L'Hopital's Rule on the fraction: \[\lim _{x \rightarrow 0} \frac{\frac{-1}{x}}{\cos (x)}\] Now we can directly evaluate this limit: \[\lim _{x \rightarrow 0} \frac{\frac{-1}{x}}{\cos (x)} = \frac{-1}{1} = -1\]
04

Reverse the natural logarithm

Since we applied the natural logarithm in step 1, we need to reverse the natural logarithm by applying the exponentiation with base \(e\): \[e^{-1} = \frac{1}{e}\]
05

State the result

The limit of the given sequence is: \[\lim _{n \rightarrow \infty} n^{\sin (\pi / n)} = \frac{1}{e}\]

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Key Concepts

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

Natural Logarithms
The natural logarithm, denoted as \( \ln(x) \), is a special type of logarithm where the base is the constant \( e \), approximately equal to 2.71828. The natural logarithm is useful in calculus and is often used to transform expressions, making them easier to manipulate. In this exercise, the term \( n^{\sin(\pi/n)} \) involves a power with a complex exponent. To simplify, we apply the natural logarithm:
  • Taking the logarithm of both sides allows us to apply the logarithm power rule: \( \ln(a^b) = b \cdot \ln(a) \).
  • This transformation simplifies our expression to \( \sin(\pi/n) \cdot \ln(n) \). This simplification is key to applying further calculus techniques effectively.
By converting to logarithmic form, we can more readily analyze and solve the limit problem. The properties of logarithms help access complex expressions, making them manageable and solvable.
L'Hopital's Rule
L'Hopital's Rule is a valuable method in calculus used to solve limits that result in indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). If a limit initially appears in one of these indeterminate forms, L'Hopital's Rule provides a way out by differentiating the numerator and denominator separately. Here鈥檚 how L'Hopital's Rule applies to our problem:
  • We needed to rewrite our limit in a form suitable for L'Hopital's Rule by expressing it as a fraction, \( \frac{\ln(n)}{\frac{1}{\sin(\pi/n)}} \).
  • As \( n \to \infty \) or equivalently, \( x \to 0 \) while we set \( x = \pi/n \), both numerator and denominator approach zero, forming a \( \frac{0}{0} \) indeterminate form.
By the rule, we differentiated \( \ln(n) \) and \( \cos(x) \), allowing us to then calculate the limit and resolve the original indeterminate form into a real number, \( -1 \), simplifying the limit evaluation.
Indeterminate Forms
Indeterminate forms occur in calculus when a limit yields an expression like \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), \( 0 \cdot \infty \), \( \infty - \infty \), among others. These forms suggest limits that cannot be directly evaluated and require special techniques to resolve. In this exercise:
  • The expression \( \lim_{x \to 0} \frac{\ln(\pi/x)}{\frac{1}{\sin(x)}} \) represents a \( \frac{0}{0} \) indeterminate form.
  • Indeterminate forms are misleading; they do not imply the limit is zero or infinite without further analysis.
Overcoming these forms involves reworking the expression, applying rules such as L'Hopital's Rule, or algebraic simplifications, all of which convert the troubling form into an evaluative one. Once resolved, the true behavior of the sequence or function near the point is revealed.
Exponential Functions
Exponential functions are characterized by expressions of the form \( a^x \), where \( a \) is a constant base and \( x \) is the exponent. They grow or decay rapidly and are pivotal in many areas of mathematics, including limits and calculus. In our sequence limit problem:
  • The original expression \( n^{\sin(\pi/n)} \) can be considered as an exponential function with base \( n \) raised to a complex exponent).
  • Complex exponents necessitate techniques such as natural logarithms for simplification, allowing for the transformation from an exponential to a linear equation.
By analyzing the limit, the exponential function's influence transforms the initial problem into one soluble by existing calculus methods. By finalizing with \( e^{-1} \), we reverberate the characteristic beauty of exponential decay and demonstrate the sequence's convergence to \( \frac{1}{e} \).

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Most popular questions from this chapter

Below we list some improper integrals. Determine whether the integral converges and, if so, evaluate the integral. $$\int_{-\infty}^{\infty} \frac{1}{e^{x}+e^{-x}} d x$$

The integral $$\int_{0}^{\infty} \frac{1}{\sqrt{x}(1+x)} d x$$ is improper in two distinct ways: the interval of integration is unbounded and the integrated is unbounded. If we rewrite the integral as $$\int_{0}^{1} \frac{1}{\sqrt{x}(1+x)} d x+\int_{1}^{\infty} \frac{1}{\sqrt{x}(1+x)} d x$$ then we have two improper integrals, the first having an unbounded integrand and the second defined on an unbounded interval. If each of these integrals converges with values \(L_{1}\) and \(L_{2}\). then the original integral converges and has value \(L_{1}+L_{2} .\) Evaluate the original integral.

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