Problem 41 - University Calculus: Early Transcendentals

Question

Use I'Hópital's rule to find the limits. $$\lim _{x \rightarrow 1^{+}}\left(\frac{1}{x-1}-\frac{1}{\ln x}\right)$$

Step-by-Step Solution

Answer

The limit is 0.

Step 1: Identify the Indeterminate Form

First, analyze the limit expression \( \lim_{x \rightarrow 1^+} \left( \frac{1}{x-1} - \frac{1}{\ln x} \right) \). As \( x \rightarrow 1^+ \), both \( \frac{1}{x-1} \) and \( \frac{1}{\ln x} \) approach \( \pm \infty \). However, when we combine them, we get an indeterminate form of \( \infty - \infty \). To resolve this, we need a common denominator.

Step 2: Combine Fractions

Combine the fractions into a single fraction with a common denominator: \( \lim_{x \rightarrow 1^+} \left( \frac{\ln x - (x-1)}{(x-1)\ln x} \right) \). Now the limit takes the form \( \frac{0}{0} \), making it a candidate for L'Hôpital's Rule.

Step 3: Apply L'Hôpital's Rule

L'Hôpital's Rule can be applied to limits of the form \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). Differentiate the numerator and the denominator separately: the derivative of \( \ln x - (x-1) \) is \( \frac{1}{x} - 1 \), and the derivative of \( (x-1)\ln x \) is \( \ln x + 1 \).

Step 4: Re-evaluate the Limit

Re-evaluate the limit using the derivatives found: \( \lim_{x \rightarrow 1^+} \frac{\frac{1}{x} - 1}{\ln x + 1} = \lim_{x \rightarrow 1^+} \frac{\frac{1-x}{x}}{\ln x + 1} = \lim_{x \rightarrow 1^+} \frac{1-x}{x(\ln x + 1)} \). Simplify further to get \( \lim_{x \rightarrow 1^+} \frac{(1-x)(x-1)}{x} \).

Step 5: Simplify and Evaluate

Since \( (1-x)(x-1) = -(x-1)^2 \), substitute back into the limit: \( \lim_{x \rightarrow 1^+} \frac{-(x-1)^2}{x(\ln x + 1)} \). As \( x \rightarrow 1^+ \), the upper expression approaches \( 0 \) because \( (x-1)^2 \rightarrow 0 \). Therefore, the limit evaluates to \( 0 \).

Key Concepts

Limits

Limits are a fundamental concept in calculus, essential for understanding how functions behave as they approach a certain point. When we say that the limit of a function exists as it approaches a specific value, we're essentially discussing how the output of the function behaves when the input gets infinitely close to a specific point.
For example, in our exercise, we're tasked with finding the limit of the expression \( \lim_{x \rightarrow 1^{+}}\left(\frac{1}{x-1}-\frac{1}{\ln x}\right) \) as \( x \) approaches 1 from the right. This involves analyzing what happens when \( x \) gets arbitrarily close to, but is still greater than, 1. The trick is to handle or simplify the expression in such a way that we can clearly see the behavior of the function near that point.
Understanding limits also allows us to determine continuity, differentiability, and many other properties of functions. They are the gateway to exploring more complex calculus topics.

L'Hôpital's Rule

L'Hôpital's Rule offers a systematic way to deal with indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) that often appear in limit evaluations. The rule states that if a limit leads to an indeterminate form after substituting the value into the function, one can take the derivatives of the numerator and denominator separately and then try to evaluate the limit again.
In our exercise, after combining the expressions into a single fraction and obtaining \( \frac{0}{0} \), we apply L'Hôpital's Rule. By differentiating the numerator \( \ln x - (x-1) \), yielding \( \frac{1}{x} - 1 \), and the denominator \( (x-1)\ln x \), yielding \( \ln x + 1 \), we transform the original problem into a more straightforward expression.
This method often simplifies complex limit problems, as it reduces the original expression to derivatives, which are typically easier to compute and interpret.

Indeterminate Forms

Indeterminate forms arise when direct substitution in a limit results in expressions like \( \infty - \infty \), \( 0 \cdot \infty \), \( 1^{\infty} \), or the more common \( \frac{0}{0} \) and \( \frac{\infty}{\infty} \). In these cases, it's unclear what the limit should be, prompting the need for additional analysis or methods such as L'Hôpital's Rule.
In our given exercise, substituting \( x \rightarrow 1^{+} \) into \( \frac{1}{x-1} - \frac{1}{\ln x} \) initially leads to the form \( \infty - \infty \). This is a classic indeterminate form because it suggests two conflicting behaviors: one part of the expression suggests an infinitely large value, while the other does the same, yet subtraction complicates resolution through direct calculation.
By converting the original expression into a single fraction, the indeterminate form \( \frac{0}{0} \) emerged, which is then aptly handled using L'Hôpital's Rule, leading to a simpler scenario for limit evaluation.