Problem 61 - Thomas Calculus in SI Units

Question

Find the limits $$\lim _{x \rightarrow \infty}\left(\frac{x+2}{x-1}\right)^{x}$$

Step-by-Step Solution

Answer

The limit is \( e^3 \).

Step 1: Rewrite the Expression

We start with the expression \( y = \left(\frac{x+2}{x-1}\right)^x \). For ease of handling the limit, we'll take the natural logarithm on both sides: \[ \ln y = x \ln \left(\frac{x+2}{x-1}\right) \]

Step 2: Simplify the Logarithmic Expression

Use the properties of logarithms. We have \[ \ln \left(\frac{x+2}{x-1}\right) = \ln(x+2) - \ln(x-1) \] Expand using first-order Taylor expansion for \(\ln(1 + u)\) where \(u\) is small:\[ \ln(x+2) - \ln(x-1) = \ln\left(1 + \frac{3}{x-1}\right) \approx \frac{3}{x-1} \]

Step 3: Substitute and Simplify

Substitute back into the logarithmic expression:\[ \ln y = x \cdot \frac{3}{x-1} \]Simplify the expression as \( x \to \infty \):\[ \ln y = \frac{3x}{x-1} = \frac{3x}{x}\frac{1}{1-\frac{1}{x}} \approx 3 \cdot 1 = 3 \]

Step 4: Apply Exponential

Since \( \ln y \approx 3 \) as \( x \to \infty \), this implies:\[ y = e^3 \]

Step 5: Conclude the Limit

Conclude by stating that \( \lim_{x \to \infty} \left(\frac{x+2}{x-1}\right)^{x} = e^3 \).

Key Concepts

Natural Logarithms

Natural logarithms are a key component in calculus, providing a way to simplify complex equations. They use the base of the natural exponential function, denoted as \( e \), where \( e \) is an irrational constant approximately equal to 2.71828. When you take the natural logarithm of a number, you are effectively finding the power to which \( e \) must be raised to arrive at that number.

• Natural logarithms can transform multiplicative relationships into additive relationships, which is often useful for solving limits involving powers and fractions.
• This property is why we initially rewrote the original problem in terms of natural logarithms, simplifying the exponential expression into a more manageable form.

Taking the natural log of \( y \) transformed the problem into an equation that was easier to handle analytically.

Taylor Expansion

The Taylor expansion is a powerful tool in approximating functions using polynomials. For a function \( f(x) \), its Taylor series expansion about a point \( a \) gives us a polynomial whose value at \( x \) is approximately that of \( f(x) \).

• In our specific solution, we used the first-order Taylor expansion: \( \ln(1 + u) \approx u \), when \( u \) is small.
• This approximation is crucial when simplifying logarithmic expressions, especially when the terms \( \frac{3}{x-1} \) become very small as \( x \) approaches infinity.

Taylor expansions help break down overall complex functions into smaller, more understandable parts by providing polynomial approximations.

Exponential Functions

Exponential functions, particularly those with the base \( e \), play a vital role in calculus and higher mathematics. The function \( e^x \) is unique because its rate of growth is proportional to its value, a property that makes it incredibly useful for modeling growth processes in nature and finance.

• After simplifying our limit problem using logarithms, we needed to "undo" the logarithm by applying the exponential function, reverting \( \ln y = 3 \) to \( y = e^3 \).
• This step emphasizes the inverse relationship between natural logarithms and exponential functions, a concept that is foundational in solving limits involving exponential terms.

Exponential functions serve as the "undoing" mechanism for logarithms, ensuring our final results are expressed in original terms.

Asymptotic Behavior

Asymptotic behavior describes how functions behave as inputs become very large (or very small), which is central to understanding limits. When seeking the limit of \( \left( \frac{x+2}{x-1} \right)^x \), we were interested in how this expression behaves as \( x \to \infty \).

• The simplification process, involving rewriting and expanding, ensures that we capture the underlying trend or behavior of the function when \( x \) is large.
• For instance, as \( x \) grows, expressions like \( \frac{1}{x} \) contribute less to the overall behavior, allowing us to approximate more simply.

Understanding asymptotic behavior lets us make accurate predictions and formulate expressions for limits, capturing the essence of what happens to the function at extremes.