Problem 51 - Thomas Calculus

Question

Find the limits in Exercises \(51-66\) $$\lim _{x \rightarrow 1^{+}} x^{1 /(1-x)}$$

Step-by-Step Solution

Answer

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

Step 1: Understand the Limit Expression

We are given the expression \( \lim_{x \to 1^+} x^{1/(1-x)} \). Our goal is to determine what value this expression approaches as \( x \) approaches 1 from the right (\( 1^+ \)). This means we'll consider values of \( x \) that are slightly greater than 1.

Step 2: Analyze the Inner Function

The exponent is \( \frac{1}{1-x} \). As \( x \to 1^+ \), \( 1-x \to 0^- \). Thus, \( \frac{1}{1-x} \to -\infty \) as we divide 1 by a negative number approaching 0. This results in the exponent becoming very large and negative.

Step 3: Consider the Base of the Exponent

Now, observe that the base of the exponent is \( x \). As \( x \to 1^+ \), \( x \to 1^+ \) (just over 1). The base gets infinitesimally close to 1.

Step 4: Rewrite the Expression

We can rewrite the original expression as \( e^{\ln(x^{1/(1-x)})} = e^{\frac{\ln x}{1-x}} \). Investigate the behavior of \( \frac{\ln x}{1-x} \) as \( x \to 1^+ \). The goal is to see what limit this expression approaches.

Step 5: Use L'Hôpital's Rule

Since the direct substitution of \( x = 1 \) in \( \frac{\ln x}{1-x} \) gives \( \frac{0}{0} \), apply L'Hôpital's Rule. Differentiate the numerator and denominator:- \( (\ln x)' = \frac{1}{x} \)- \( (1-x)' = -1 \)Thus, the limit becomes \( \lim_{x \to 1^+} \frac{1/x}{-1} = \lim_{x \to 1^+} -\frac{1}{x} = -1 \).

Step 6: Finalize the Limit Expression

Since \( \frac{\ln x}{1-x} \to -1 \) as \( x \to 1^+ \), it follows that \( e^{\frac{\ln x}{1-x}} \to e^{-1} = \frac{1}{e} \). Thus, the original limit expression evaluates to \( \frac{1}{e} \).

Key Concepts

L'Hôpital's Rule

L'Hôpital's Rule is a powerful mathematical tool used to evaluate limits that initially appear to result in indeterminate forms, such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). The rule states that if you encounter these forms, you can take the derivative of the numerator and the derivative of the denominator separately, then re-evaluate the limit.
This approach often simplifies the expression, making it easier to find the limit.
In our exercise with \( \lim_{x \to 1^+} x^{1/(1-x)} \), we rewrote the problem as \( \frac{\ln x}{1-x} \) and noticed it was in the \( \frac{0}{0} \) form when substituting \( x = 1 \).

• By applying L'Hôpital's Rule, we differentiated \( \ln x \) to get \( \frac{1}{x} \) and \( 1-x \) to get \(-1\).
• Then we re-calculated the limit as \( \lim_{x \to 1^+} -\frac{1}{x} \), which simplified to \(-1\).

This result was used to determine the behavior of the initial expression, ultimately showing that it approaches \( \frac{1}{e} \). Understanding when and how to use L'Hôpital's Rule effectively can greatly assist in evaluating complex limits.

Exponentiation in Limits

When dealing with limits that involve exponents, such as \( x^{1/(1-x)} \) in our problem, it's helpful to think about how both the base and the exponent behave as the variable approaches a particular value.
As seen in the exercise, the exponent \( \frac{1}{1-x} \) becomes very large and negative as \( x \to 1^+ \) because \( 1-x \) approaches \( 0^- \).
Meanwhile, the base \( x \) approaches \( 1^+ \).

• One effective technique when dealing with such scenarios is to apply the natural logarithm (ln) in order to transform the expression. This allows for potentially using L'Hôpital's Rule or other simplifications.
• Rewriting the problem as \( e^{\ln(x^{1/(1-x)})} = e^{\frac{\ln x}{1-x}} \) transforms the expression into a form where the denominator and numerator can be analyzed more easily.

The new challenge becomes finding \( \lim_{x \to 1^+} \frac{\ln x}{1-x} \), which is more straightforward than directly computing the limit of an exponentiating function.

Ln (Natural Logarithm)

The natural logarithm, denoted as \( \ln \), is a fundamental tool in calculus, particularly when working with exponential functions and limits. \( \ln x \) is the logarithm of \( x \) to the base \( e \), where \( e \) is approximately 2.71828.
In our problem, we use \( \ln \) to "bring down" an exponent in the expression \( x^{1/(1-x)} \), making it more manageable to apply techniques like L'Hôpital's Rule.
By converting \( x^{1/(1-x)} \) to \( e^{\frac{\ln x}{1-x}} \), we've essentially turned a complex exponent problem into an issue of limits and derivatives.

• This transformation is beneficial because \( \ln \) functions are continuous and differentiable, making them suitable for calculus operations.
• The application of \( \ln \) in this context demonstrates its usefulness in simplifying and solving indeterminate form limits.
• Understanding this process is crucial, as \( \ln \) serves not just to handle exponents but to make otherwise difficult calculus problems more approachable.

Mathematically savvy usage of \( \ln \) allows us to navigate through the complexities of limit problems such as those involving both exponents and L'Hôpital's Rule.