Problem 62 - University Calculus: Early Transcendentals

Question

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

Step-by-Step Solution

Answer

The limit is 1.

Step 1: Identify the Form

As we examine the expression \( \left(\frac{x^{2}+1}{x+2}\right)^{1/x} \), we recognize that when \( x \to \infty \), the base \( \frac{x^{2}+1}{x+2} \) approaches \( x \) as the quadratic term dominates. This produces the undefined form \( 1^{\infty} \), suggesting the use of the exponential limit approach.

Step 2: Use the Natural Logarithm

To handle the \( 1^{\infty} \) form, take the natural logarithm: \[ \ln L = \lim_{x \to \infty} \frac{1}{x} \ln \left( \frac{x^2 + 1}{x + 2} \right). \] Simplifying the argument of the logarithm, this becomes: \[ \ln \left( \frac{x^2 + 1}{x + 2} \right) = \ln(x^2 + 1) - \ln(x + 2). \]

Step 3: Simplify the Logarithmic Difference

We now analyze each component:- \( \ln(x^2 + 1) = \ln(x^2(1 + 1/x^2)) = 2\ln(x) + \ln(1 + 1/x^2) \)- \( \ln(x + 2) = \ln(x(1 + 2/x)) = \ln(x) + \ln(1 + 2/x) \)Thus, the expression becomes \[ \ln L = \lim_{x \to \infty} \frac{1}{x} \left( (2\ln(x) + \ln(1 + 1/x^2)) - (\ln(x) + \ln(1 + 2/x)) \right). \]

Step 4: Combine and Simplify Further

Notice that \( 2\ln(x) - \ln(x) = \ln(x) \), so we simplify:\[ \ln L = \lim_{x \to \infty} \frac{1}{x} \left( \ln(x) + \ln(1 + 1/x^2) - \ln(1 + 2/x) \right). \] Break down further:\[ \ln L = \lim_{x \to \infty} \left( \frac{\ln(x)}{x} + \frac{\ln(1 + 1/x^2)}{x} - \frac{\ln(1 + 2/x)}{x} \right). \]

Step 5: Evaluate Individual Limits

Evaluate each term separately:- For \( \frac{\ln(x)}{x} \), it approaches 0 as \( \ln(x) \) grows slower than \( x \).- For \( \frac{\ln(1 + 1/x^2)}{x} \), as \( x \to \infty \), this approaches 0 because \( \ln(1 + 1/x^2) \) approaches 0 faster.- For \( \frac{\ln(1 + 2/x)}{x} \), it simplifies to \( \frac{2/x}{x} = \frac{2}{x^2} \) which approaches 0.Thus, \( \ln L = 0 \).

Step 6: Calculate the Limit

Since \( \ln L = 0 \), the limit \( L \) itself must be:\[ L = e^{0} = 1. \] So, the initial limit expression evaluates to 1 as \( x \to \infty \).

Key Concepts

Logarithmic Differentiation

Logarithmic differentiation is a powerful technique in calculus used to differentiate functions that are products, quotients, or powers more easily. This method involves taking the natural logarithm of both sides of an equation. By doing so, we often transform a complex expression into a simpler one. Here’s why logarithmic differentiation can be useful:

• It simplifies the differentiation process, especially for functions like products and quotients.
• It turns multiplication into addition, making it easier to differentiate.

Let's say we have a function that is a power like \[ y = u^{v} \] By taking the natural logarithm, we transform this into \[ \ln y = v \ln u \]. This allows us to differentiate using the chain rule. The practical beauty of this technique is evident in cases like the one seen in the original exercise, where we deal with a limit scenario involving exponential expressions.

Exponential Limits

Exponential limits are crucial when dealing with expressions that have variables both in the base and the exponent. A typical case involves limits of the form \[ ext{Base}^{ ext{Exponent}} \] as variable x approaches infinity. In such scenarios, you might encounter an indeterminate form such as \(1^{rac{1}{x}}\) or \(0^{rac{1}{x}}\).Taking the natural logarithm helps unravel these complex cases by using the 1. power property of logarithms, that is \( ext{log}(a^{b}) = b \times ext{log}(a) \).2. properties that change the original form \( Base^{ ext{Exponent}} \) into a more manageable linear expression.Once the limit is simplified and evaluated, you must exponentiate the final result. This generates the real value of the expression, eliminating the confusion spurred by the original indeterminate form.

Indeterminate Forms

In calculus, indeterminate forms like \( \frac{0}{0} \), \(\infty - \infty \), and \(1^{\infty}\) arise during limit evaluation. These forms are termed indeterminate because their actual value is not directly apparent and further analysis is required.The form \(1^{\infty}\) shows up often in exponential limits when the base approaches 1 and the exponent grows without bound, as seen in the solved exercise.Here are general approaches to handle indeterminate forms:

• Use the natural logarithm to simplify the expression (as with logarithmic differentiation).
• Transform the expression algebraically into a form suitable for known limit laws or L'Hôpital's Rule.

Understanding these anomalies and knowing how to resolve them using appropriate calculus techniques is crucial for accurate evaluation of limits.