Problem 56 - Single Variable Calculus

Question

Evaluate the limit using l'Hôpital's Rule if appropriate. $$ \lim _{x \rightarrow \infty}\left(a^{1 / x}-1\right) x $$

Step-by-Step Solution

Answer

The short answer is: \(\lim_{x \rightarrow \infty}\left(a^{1/x}-1\right)x = \ln a.\)

Step 1: Rewrite the expression

We will rewrite the given expression into a more convenient form that can be solved using l'Hôpital's Rule. The expression is: \[ \lim_{x\rightarrow\infty} (a^{1/x}-1)x. \] Let's consider a substitution: \[ y = \frac{1}{x}. \] This implies that as \(x\rightarrow\infty\), \(y\rightarrow0\), and our expression becomes: \[ \lim_{y \rightarrow 0}\frac{a^y - 1}{y}. \] Now, the expression is in a more convenient form for applying l'Hôpital's Rule, since it is in the form of an indeterminate type \(\frac{0}{0}\).

Step 2: Apply l'Hôpital's Rule

Since our limit is now in the form \(\frac{0}{0}\), we can apply l'Hôpital's Rule. To do this, we need to differentiate the numerator and the denominator with respect to y: \[ \frac{d}{dy}(a^y-1) = a^y \ln a, \] and \[ \frac{d}{dy}(y) = 1. \] Now, apply l'Hôpital's Rule: \[ \lim_{y \rightarrow 0}\frac{a^y-1}{y} = \lim_{y \rightarrow 0}\frac{a^y \ln a}{1}. \]

Step 3: Evaluate the limit

Taking the limit as \(y\rightarrow0\) in the expression obtained after applying l'Hôpital's Rule: \[ \lim_{y \rightarrow 0}\frac{a^y \ln a}{1} = \frac{a^0 \ln a}{1} = \frac{\ln a}{1} = \ln a. \] Our final answer is: \[ \lim_{x \rightarrow \infty}\left(a^{1/x}-1\right)x = \ln a. \]

Key Concepts

Indeterminate Forms

In calculus, indeterminate forms often occur when evaluating limits. They are expressions where enough information isn't available to determine the result directly. A common type is \( \frac{0}{0} \), seen when the limit of the numerator and the denominator both approach zero.

These forms are essential because they indicate that further analysis is needed to find an accurate solution.

• Examples include \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), \( \infty - \infty \).
• The techniques such as l'Hôpital's Rule help resolve these forms by differentiating the involved functions.
• Identifying an indeterminate form is the first step in applying a remedy like l'Hôpital's Rule to evaluate limits.

Understanding these forms is important as they often appear in more complex mathematical problems.

Differentiation

Differentiation is a mathematical process used to find how a function changes as its input changes. In the context of l'Hôpital's Rule, differentiating the numerator and denominator separately can simplify solving a limit problem.

The derivative gives us the rate of change or the slope of the function at a given point.

• The derivative of \( a^y \) with respect to \( y \) is \( a^y \ln(a) \).
• The derivative of \( y \) with respect to itself is simply 1.
• By applying these derivatives, we transform the original limit problem into a simpler form.

Differentiation is vital in calculus as it allows us to understand and solve problems dealing with change and trends effectively.

Limits

A limit helps us understand the behavior of a function as it approaches a particular input value. It is a fundamental concept in calculus and essential for topics like continuity, derivatives, and integrals.

When evaluating limits, especially those involving indeterminate forms, strategies like substitution can be useful.

• For example, substituting \( y = \frac{1}{x} \) makes the limit easier to analyze.
• With this substitution, the limit of the original function \( \lim_{x\rightarrow\infty} \) can be explored as \( x \) goes to infinity.
• The substitution also shifts the limit to \( y \rightarrow 0 \), making it a more straightforward expression to handle.

Limits are about predicting outcomes and understanding precision, which are crucial skills in higher mathematics.

Exponential Functions

Exponential functions are an important class of functions in mathematics, characterized by constant base numbers raised to variable exponents. They appear frequently in calculus due to their unique growth properties.

These functions are particularly interesting as they model growth over continuous time and are used in various real-world applications.

• The general form is \( a^x \), where \( a \) is a constant base.
• When differentiating an exponential function \( a^y \), where \( y \) is the exponent, a logarithmic factor \( \ln(a) \) appears.
• In our original problem, the function \( a^{1/x} \) represents an inverse exponential scenario, crucial for transforming it into an indeterminate form.

Understanding exponential functions is fundamental because of their wide applicability and role in natural phenomena modeling.