Problem 32 - Calculus Early Transcendentals: Pearson New International Edition

Question

Be sure you have an indeterminate form before applying l'Hôpital's Rule. $$\lim _{x \rightarrow 1}\left(\frac{1}{x-1}-\frac{x}{\ln x}\right)$$

Step-by-Step Solution

Answer

The limit is 0.

Step 1: Identify Indeterminate Form

Substitute \( x = 1 \) into the expression \( \frac{1}{x-1} - \frac{x}{\ln x} \). We get \( \frac{1}{0} - \frac{1}{\ln(1)} = \frac{1}{0} - \frac{1}{0} \), which is of the form \( \infty - \infty \). This is an indeterminate form that can be transformed for l'Hôpital's Rule.

Step 2: Common Denominator

To use l'Hôpital's Rule, first write the expression with a common denominator: \( \frac{\ln x - x(x-1)}{(x-1)\ln x} \). Simplify the numerator to \( \ln x - (x^2 - x) = \ln x - x^2 + x \).

Step 3: Check Indeterminate Form Again

Evaluate the limit by substituting \( x = 1 \) into \( \frac{\ln x - x^2 + x}{(x-1)\ln x} \). We find \( \frac{0 - 1^2 + 1}{0 \cdot 0} = \frac{0}{0} \), confirming another indeterminate form suitable for l'Hôpital's Rule.

Step 4: Apply l'Hôpital's Rule

Differentiate numerator and denominator separately: **Numerator**: \( \frac{d}{dx}(\ln x - x^2 + x) = \frac{1}{x} - 2x + 1 \), **Denominator**: \( \frac{d}{dx}((x-1)\ln x) = \ln x + \frac{x-1}{x} \).

Step 5: Simplify and Evaluate Limit

Substitute \( x = 1 \) into the new expression \( \frac{\frac{1}{x} - 2x + 1}{\ln x + \frac{x-1}{x}} \): The expression becomes \( \frac{1/1 - 2(1) + 1}{0 + 0} = \frac{1 - 2 + 1}{1} = 0 \). Since the limit resolves to a finite value, it evaluates to 0.

Key Concepts

Indeterminate Forms

When dealing with limits, we often encounter expressions that cannot be easily evaluated at first glance. These expressions are known as "indeterminate forms."
They typically arise in situations where direct substitution results in ambiguous expressions like \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), or in our case, \( \infty - \infty \). Indeterminate forms signal that we need to apply special techniques to find the limit.

• For \( \frac{0}{0} \), we must analyze the behavior of the functions involved. Often, rewriting or factoring is needed.
• In cases like \( \infty - \infty \), we need a strategy to combine terms efficiently, turning them into a form where we can apply rules like l'Hôpital's Rule.

Recognizing these forms is crucial because it tells us that we cannot determine the limit without further manipulation. In the example problem, substituting \( x = 1 \) results in \( \frac{1}{0} - \frac{1}{0} \), which is \( \infty - \infty \). This identification is the first important step in problem-solving.

Common Denominator

A common denominator is a shared multiple of the denominators of several fractions. When solving limits involving multiple fractions, finding a common denominator is often necessary. This allows us to combine the fractions into a single fraction, making it easier to manipulate and apply techniques like l'Hôpital's Rule.
Let's consider the expression \( \frac{1}{x-1} - \frac{x}{\ln x} \). Initially, these fractions have different denominators, \( x-1 \) and \( \ln x \), requiring us to find a common one to make them into a single fraction.
To do this, we multiply both fractions by the other's denominator, giving us:

• \( \frac{\ln x}{(x-1)\ln x} \)
• \( \frac{x(x-1)}{(x-1)\ln x} \)

Combining these gives \( \frac{\ln x - x(x-1)}{(x-1)\ln x} \). This simplification is crucial for further evaluation, as it helps us to identify the limit's indeterminate form and apply more sophisticated tools to solve it.

Limits in Calculus

Limits are a fundamental concept in calculus, allowing us to understand the behavior of functions as they approach specific values. They form the basis for derivatives and integrals, both of which are central to calculus.
Finding the limit of a function involves determining the value that the function approaches as the variable gets close to a specified point. Sometimes, this is straightforward, but often it requires special methods, especially when dealing with indeterminate forms.
For example, in our problem, the limit \[ \lim_{x \to 1} \left(\frac{\ln x - x^2 + x}{(x-1)\ln x}\right) \] is initially in the \( \frac{0}{0} \) form after simplification. Since direct evaluation is not possible, techniques like l'Hôpital's Rule become extremely useful. This rule states that if the limits of the top and bottom derivatives exist, then the original limit can be found by taking these derivatives.
It is this interplay between limits and their properties that leads to deeper insights in calculus, allowing us to solve problems and understand the change.