Problem 60 - Calculus for Biology and Medicine

Question

Find the limits in Problems 1-60; not all limits require use of l'Hôpital's rule. $$ \lim _{x \rightarrow \infty}\left(\frac{x+1}{x+2}\right)^{x} $$

Step-by-Step Solution

Answer

The limit is \( \frac{1}{e} \).

Step 1: Identify the form of the limit

We need to find \( \lim _{x \rightarrow \infty}\left(\frac{x+1}{x+2}\right)^{x} \). As \( x \rightarrow \infty \), the base of the exponential approaches 1. This results in an indeterminate form \( 1^\infty \).

Step 2: Take the natural logarithm

For limits involving \( 1^\infty \), a common technique is to set \( y = \left(\frac{x+1}{x+2}\right)^{x} \) and consider \( \ln y = x \ln \left(\frac{x+1}{x+2}\right) \). Thus, we will find \( \lim_{x \rightarrow \infty} \ln y \).

Step 3: Simplify \( \ln \left(\frac{x+1}{x+2}\right) \)

Rewrite \( \ln \left(\frac{x+1}{x+2}\right) \) using the properties of logarithms: \( \ln \left(\frac{x+1}{x+2}\right) = \ln(x+1) - \ln(x+2) \).

Step 4: Use series expansion for logarithms

Using the identity \( \ln(1 + u) \approx u \) for small \( u \), we have\[\ln(x+1) - \ln (x+2) \approx \ln \left(1 + \frac{1}{x} \right) - \ln \left(1 + \frac{2}{x} \right) \approx \frac{1}{x} - \frac{2}{x},\]which simplifies to \( -\frac{1}{x} \).

Step 5: Evaluate the limit of the product

Substitute \( \ln \left(\frac{x+1}{x+2}\right) \approx -\frac{1}{x} \) into the expression for \( \ln y \), thus:\[ \lim_{x \rightarrow \infty} x \ln \left(\frac{x+1}{x+2}\right) = \lim_{x \rightarrow \infty} x\left(-\frac{1}{x}\right) = \lim_{x \rightarrow \infty} -1 = -1.\]

Step 6: Conclude by exponentiating the result

Since \( \ln y \to -1 \), exponentiate both sides to find \( y \):\[y = e^{-1} = \frac{1}{e}.\]Thus, the original limit evaluates to \( \frac{1}{e} \).

Key Concepts

Exponential Indeterminate Forms

In calculus, limits often present themselves in indeterminate forms, meaning that their ultimate behavior is not immediately clear. One of the classic indeterminate forms is \(1^\infty\). This form arises when you have an expression raised to the power of, and itself approaching, infinity. At first glance, it might seem straightforward, but the outcome can be quite unexpected due to the balance between a base approaching 1 and an exponent rapidly growing. Indeterminate forms require special techniques to resolve and determine their exact limit.

• Indeterminate forms are situations in calculus where standard limits rules don't directly provide the answer.
• In the case of \(1^\infty\), the base approaches 1, but with the power growing infinitely large, its effect can't be ignored.
• This form often requires manipulation, either through algebraic simplification, use of natural logarithms, or L'Hôpital's Rule, to find a determinate limit.

Understanding how to handle such forms is crucial for deeper calculus studies, as it opens the door to analyzing many more complex functions and their behaviors.

Natural Logarithms

Natural logarithms, denoted as \(\ln\), are logarithms with the base \(e\), an irrational constant approximately equal to 2.71828. They play a fundamental role in calculus, particularly when dealing with growth and decay problems, and help simplify complex exponential expressions, especially when dealing with indeterminate forms.

• The natural logarithm has the convenient property that it converts multiplication into addition, simplifying the manipulation of equations.
• It is particularly useful for transforming indeterminate forms into forms where limits can be more easily applied.
• In the solution of exponential limits like the one in our problem, taking the natural logarithm is a crucial step because it helps isolate the exponent, making it easier to analyze.

The key idea when using natural logarithms in limits is to work with properties like \(\ln(ab) = \ln(a) + \ln(b)\) and \(\ln(a^b) = b\ln(a)\), which facilitate breaking down the problem into simpler, more manageable parts.

Series Expansion for Logarithms

Series expansion for logarithms is a powerful tool used in calculus to approximate natural logarithms for small values of \(u\). If \(u\) is small, then \(\ln(1 + u)\) can be approximated by the series \(u - \frac{u^2}{2} + \frac{u^3}{3} - \ldots\), but for very small \(u\), we often simplify this to \(u\). This technique is particularly useful in evaluating limits by simplifying the expressions.

• Series expansion offers a method to approximate functions, helping convert complex expressions into simple, tractable forms.
• In the limit problem we're dealing with, \(\ln(1 + \frac{1}{x})\) simplifies to \(\frac{1}{x}\) because as \(x\) approaches infinity, \(\frac{1}{x}\) becomes very close to zero.
• This approximation is key for calculating limits when direct substitution leads to indeterminate forms.

Using series expansions for logarithms provides a straightforward and elegant way to handle tricky limit problems, where direct computation isn't feasible. It transforms a complex operation into a simple subtraction or division, making it possible to evaluate the limit directly.