Problem 61 - University Calculus: Early Transcendentals

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: Understand the Limit Form

The limit given is in the form \( \left(\frac{x+2}{x-1}\right)^{x} \) as \( x \to \infty \). This is an indeterminate form \( 1^{\infty} \), as both the base and the exponent drive it to an undefined state. We need to transform this expression to find the limit.

Step 2: Transform Using Natural Logarithm

To simplify the power expression, take the natural logarithm: \( y = \left(\frac{x+2}{x-1}\right)^x \). Then, \( \ln(y) = x \cdot \ln\left(\frac{x+2}{x-1}\right) \). We'll first find \( \lim_{x \to \infty} x \cdot \ln\left(\frac{x+2}{x-1}\right) \) and exponentiate the result later.

Step 3: Simplify the Logarithmic Expression

Express \( \ln\left(\frac{x+2}{x-1}\right) \) using the difference of logarithms: \( \ln(x+2) - \ln(x-1) \).For large \( x \), use the approximation \( \ln(a + b) \approx \ln(a) + \frac{b}{a} \). Hence, \( \ln(x+2) \approx \ln(x) + \frac{2}{x} \) and \( \ln(x-1) \approx \ln(x) - \frac{1}{x} \). Consequently,\[\ln\left(\frac{x+2}{x-1}\right) \approx \frac{2}{x} + \frac{1}{x} = \frac{3}{x}.\]

Step 4: Evaluate the Limit of the Logarithmic Expression

Substitute the simplified expression back into the original limit condition: \( x \cdot \ln\left(\frac{x+2}{x-1}\right) \approx x \cdot \frac{3}{x} = 3 \).Therefore, \( \lim_{x \to \infty} x \cdot \ln\left(\frac{x+2}{x-1}\right) = 3 \).

Step 5: Find the Original Limit

Since \( \ln(y) \to 3 \) as \( x \to \infty \), the expression \( y \) must approach \( e^3 \).Thus, \( \lim _{x \rightarrow \infty}\left(\frac{x+2}{x-1}\right)^{x} = e^3 \).

Key Concepts

Limits

The concept of limits is fundamental in calculus. We use limits to describe the behavior of a function as the input approaches a certain value. In this exercise, we are interested in the behavior of the expression \( \left(\frac{x+2}{x-1}\right)^x \) as \( x \to \infty \).

• When \( x \to \infty \), the limits can reveal long-term trends of functions.
• Limits help determine the value a function approaches but may not actually reach.

In our case, the function seems complex due to its form. However, by applying limits, we can uncover the simple, elegant outcome it approaches at infinity. Understanding how functions behave when inputs become very large or very small is a powerful tool in calculus.

Indeterminate Forms

Indeterminate forms occur when the limits of two functions interacting take on undefined values. They can appear in many forms, such as \( 0/0 \), \( \infty/\infty \), or as in our case, \( 1^\infty \).

• An indeterminate form means that direct substitution in a limit won't provide a clear answer.
• These forms signal the need for algebraic manipulation or calculus techniques to find the limit.

In the example given, \( \left(\frac{x+2}{x-1}\right)^x \) as \( x \to \infty \), represents a \( 1^\infty \) form. This forms a basis for more work to be done to derive a meaningful result. By manipulating the expression, often through logarithms or series expansion, we can resolve these indeterminate expressions into a tangible limit.

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is a vital tool used to simplify expressions, especially in calculus. It turns exponential problems into more manageable linear forms.

• \( \ln \) has the base \( e \), an important mathematical constant approximately 2.71828.
• It is particularly useful for transforming complex exponentials.

In the given solution, \( y = \left(\frac{x+2}{x-1}\right)^x \) is transformed into \( \ln(y) = x \cdot \ln\left(\frac{x+2}{x-1}\right) \) to tractably handle the power term. This usage allows a difficult exponent to be dealt with as a simpler multiplication, revealing the behavior of the function as \( x \to \infty \) and leading to a successful determination of the limit.

Exponential Functions

Exponential functions involve a constant base raised to a variable exponent. They're common in many fields due to their growth properties. In the context of limits, understanding these functions helps us decode rapidly growing or shrinking scenarios.

• Exponential growth illustrates increase at an accelerating rate.
• When transformed back using logarithms, they can clarify complex limits.

In the exercise, we eventually find \( y = e^3 \) as \( x \rightarrow \infty \) after resolving the expression using natural logs. This outcome indicates a function reaching a stable exponential growth, an example of how logarithmic and exponential functions interplay to simplify limit computation.