Problem 51 - University Calculus: Early Transcendentals

Question

Find the limits. $$\lim _{x \rightarrow 1^{-}} x^{1 /(1-x)}$$

Step-by-Step Solution

Answer

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

Step 1: Understand the Problem

We need to find the limit of the expression \(x^{1/(1-x)}\) as \(x\) approaches 1 from the left. This means that we are considering values of \(x\) that are slightly less than 1.

Step 2: Transform the Expression Using Logarithms

Since the expression involves an exponent that is not straightforward to evaluate, we take the natural logarithm to simplify the limit.\[y = x^{1/(1-x)} \\ln y = \frac{\ln x}{1-x}\]Thus, we now have to evaluate \(\lim_{x \to 1^-} \frac{\ln x}{1-x}\).

Step 3: Evaluate the Limit Using L'Hôpital's Rule

The expression \(\frac{\ln x}{1-x}\) becomes \(\frac{0}{0}\) as \(x\) approaches 1 from the left. This allows us to apply L'Hôpital's Rule, which states that:\[\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)},\]if the limit exists. In our case:- \(f(x) = \ln x\) and \(f'(x) = \frac{1}{x}\)- \(g(x) = 1-x\) and \(g'(x) = -1\).

Step 4: Apply L'Hôpital's Rule

Apply L'Hôpital's Rule to the limit:\[\lim_{x \to 1^-} \frac{\ln x}{1-x} = \lim_{x \to 1^-} \frac{1/x}{-1} = -\lim_{x \to 1^-} \frac{1}{x} = -1\]So, the limit of \(\ln y\) as \(x\) approaches 1 from the left is \(-1\).

Step 5: Exponentiate to Find the Original Limit

Since \(\ln y\) approaches \(-1\), we rewrite \(y\) as:\[y = e^{\ln y} \y = e^{-1} = \frac{1}{e}\]Thus, the original limit is \(\frac{1}{e}\).

Key Concepts

L'Hôpital's Rule

L'Hôpital's Rule is a valuable tool in calculus for finding limits that initially result in an indeterminate form, like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). This rule allows us to differentiate the numerator and the denominator separately.
This is particularly useful when direct substitution into the limit gives an indeterminate form. For example, in the exercise, after taking the natural logarithm and reformulating the limit problem to \(\lim_{x \to 1^-} \frac{\ln x}{1-x}\), we get \(\frac{0}{0}\) as \(x\) approaches 1 from the left.
Here's when L'Hôpital's Rule becomes handy:

• Differentiation of the numerator: Since \(f(x) = \ln x\), the derivative \(f'(x) = \frac{1}{x}\).
• Differentiation of the denominator: With \(g(x) = 1-x\), the derivative \(g'(x) = -1\).

We substitute these derivatives back into the original limit problem:\[\lim_{x \to 1^-} \frac{\ln x}{1-x} = \lim_{x \to 1^-} \frac{\frac{1}{x}}{-1} = -\lim_{x \to 1^-} \frac{1}{x} = -1. \]By converting the indeterminate limit problem to a more manageable form, L'Hôpital's Rule helps students navigate complex calculus problems efficiently.

Natural Logarithms

Natural logarithms have a base of \(e\), an irrational and transcendental number approximately equal to 2.71828. They are used to solve equations or simplify expressions involving exponents by transforming multiplicative relationships into additive ones.
In the given exercise, we used the natural logarithm to simplify the evaluation of the limit. By representing \(x^{1/(1-x)}\) as \(e^{\ln(x^{1/(1-x)})}\), we transformed the problem into:\[\ln y = \frac{\ln x}{1-x}.\]Using properties of logarithms like \(\ln(a^b) = b \ln a\), complicated power functions can be converted into simpler forms. This simplification was crucial because it allowed us to eventually apply L'Hôpital's Rule.
Understanding natural logarithms is vital in calculus because they frequently appear in problems involving rates of growth and decay, such as exponential functions. This concept is key when dealing with compound interest, population models, and more, highlighting their importance beyond textbook exercises.

Exponential Functions

Exponential functions feature the mathematical constant \(e\) and are crucial in modeling growth or decay processes.
The basic form \(y = e^x\) showcases constant and rapid growth as \(x\) increases. They appear often in real-world scenarios, such as modeling population growth, radioactive decay, and bank interest calculations.
In the context of this exercise, the expression \(x^{1/(1-x)}\) translated into \(e^{\ln(x^{1/(1-x)})}\) involves exponential functions because \(e\) becomes the base of the transformed expression.

• The central step involves evaluating \(y = e^{\ln y}\), where the property \(a^{\ln b} = b^{\ln a}\) aids the transformation.
• After simplifying, the exponential function emerges strongly: \(y = e^{-1} = \frac{1}{e}\).

This concept is crucial because exponential functions model various phenomena including:

• Biological growth patterns
• Financial investment growth
• Charge and discharge in electronics

Understanding exponential functions and their properties aids students in both academically challenging contexts and practical situations, making them an essential part of learning calculus.