Problem 29 - Calculus Single Variable

Question

Use an algebraic manipulation to put the limit in a form which can be treated using l'Hôpital's Rule; then evaluate the limit. \(\lim _{x \rightarrow 1^{+}}(x-1)^{-1} \ln (x)\)

Step-by-Step Solution

Answer

The limit evaluates to 1.

Step 1: Identify the Form

First, identify the form of the limit expression as you approach x from the right towards 1. You have \((x-1)^{{-1}}\ln(x)\). As \(x\to1^+\), \(x-1\to0^+\) and \(\ln(x)\to\ln(1)=0\). This means we have a potential \(\frac{0}{0}\) form that is suitable for l'Hôpital's Rule.

Step 2: Rewrite the Expression

To apply l'Hôpital's Rule, we need to rewrite the expression to match the \(\frac{0}{0}\) indeterminate form. Recognize that \((x-1)^{-1}\) can be rewritten as \(\frac{1}{x-1}\). Thus, the given expression becomes \(\lim_{x \rightarrow 1^{+}}\frac{\ln(x)}{x-1}\).

Step 3: Verify Indeterminate Form

Check if the rewritten limit \(\frac{\ln(x)}{x-1}\) results in a \(\frac{0}{0}\) form as \(x \rightarrow 1^{+}\). As discussed, \(\ln(1)=0\) and \(x-1=0\), confirming a \(\frac{0}{0}\) indeterminate form.

Step 4: Apply l'Hôpital's Rule

Since the expression is in the \(\frac{0}{0}\) indeterminate form, apply l'Hôpital's Rule, which involves differentiating the numerator and the denominator separately. Differentiate \(\ln(x)\) to get \(\frac{1}{x}\) and differentiate \(x-1\) to get 1. The limit becomes \(\lim_{x \rightarrow 1^{+}} \frac{\frac{1}{x}}{1}\).

Step 5: Evaluate the New Limit

Now, simplify the expression to \(\lim_{x \rightarrow 1^{+}} \frac{1}{x}\). As \(x\) approaches 1 from the right, \(\frac{1}{x}\) approaches \(1\).

Step 6: Conclusion

Therefore, the limit \(\lim _{x ightarrow 1^{+}}(x-1)^{-1}\ln(x)\) evaluates to 1 after applying l'Hôpital's Rule.

Key Concepts

Indeterminate Forms

Indeterminate forms are expressions that arise when evaluating certain limits and do not immediately indicate what the limit value is. These forms typically present as \(\frac{0}{0}\), \(\frac{\infty}{\infty}\), \(0 \times \infty\), or similar ambiguous expressions. A classic example is when both the numerator and the denominator of a fraction approach zero as x approaches a certain value. Indeterminate forms are crucial in calculus as they require further analysis or manipulation to determine what the limit actually is.

• To ascertain an indeterminate form, identify expressions that appear to be undefined or ambiguous as the variable approaches a specified value.
• Recognizing such forms allows us to apply specific mathematical strategies, like l'Hôpital's Rule, to progress with limit evaluation.

Indeterminate forms enable mathematicians to use powerful tools to explore limits that are not initially clear.

Limit Evaluation

Limit evaluation is the process of finding the value that a function approaches as the input gets arbitrarily close to a certain point. It serves as a foundational concept in calculus, enabling the analysis of functional behavior as variables tend toward particular values.

• When tackling limits, start by substituting the approaching value into the function, unless it leads to an indeterminate form.
• If direct substitution results in an indeterminate form like \(\frac{0}{0}\), additional techniques must be utilized.
• One such technique is l'Hôpital's Rule, which simplifies the evaluation process by providing a method to differentiate the numerator and denominator.

Using limit evaluation strategies such as substitution, factoring, and algebraic manipulation ensures a comprehensive understanding of how functions behave close to specific points.

Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying expressions using mathematical operations to solve problems or clarify their structure. This process is crucial when dealing with limits, especially when attempting to transform indeterminate forms into determinate ones.

• By rewriting expressions, the hidden behavior of functions as variables approach certain limits can be better comprehended.
• In the context of the provided exercise, algebraic manipulation involved expressing \( (x-1)^{-1} \) as \( \frac{1}{x-1} \), making it suitable for applying l'Hôpital's Rule.
• This transformation is key to enabling the evaluation of limits, which may not be straightforward at first glance.

Through strategic algebraic manipulation, complex limit problems become more approachable, enabling the use of advanced calculus techniques for evaluation.