Question

Find the indicated limit. Make sure that you have an indeterminate form before you apply l'Hôpital's Rule. $$ \lim _{x \rightarrow 1} \frac{\ln x^{2}}{x^{2}-1} $$

Short answer

The limit is 1.

Step-by-step solution

Identify Indeterminate Form

First, check if the limit results in an indeterminate form, such as \( \frac{0}{0} \). Substitute \( x = 1 \) into both the numerator and the denominator:- Numerator: \( \ln(1^2) = \ln 1 = 0 \)- Denominator: \( 1^2 - 1 = 0 \)Since substituting \( x = 1 \) gives \( \frac{0}{0} \), it is an indeterminate form, and we can apply l'Hôpital's Rule.

Apply l'Hôpital's Rule

Since we have an indeterminate form \( \frac{0}{0} \), apply l'Hôpital's Rule. Differentiate the numerator and the denominator:- Differentiating the numerator: \( \frac{d}{dx}(\ln x^2) = \frac{d}{dx}(2 \ln x) = \frac{2}{x} \)- Differentiating the denominator: \( \frac{d}{dx}(x^2 - 1) = 2x \)Apply the rule:\[ \lim _{x \rightarrow 1} \frac{\ln x^{2}}{x^{2}-1} = \lim _{x \rightarrow 1} \frac{\frac{2}{x}}{2x} \]

Simplify the Expression

Simplify the expression obtained after applying l'Hôpital's Rule:\[ \lim _{x \rightarrow 1} \frac{\frac{2}{x}}{2x} = \lim _{x \rightarrow 1} \frac{2}{2x^2} = \lim _{x \rightarrow 1} \frac{1}{x^2} \]

Evaluate the Limit

Substitute \( x = 1 \) into the simplified expression:\[ \lim _{x \rightarrow 1} \frac{1}{x^2} = \frac{1}{1^2} = 1 \]The limit of the function as \( x \) approaches 1 is 1.

Key concepts

Indeterminate Forms

In calculus, an indeterminate form is a mathematical expression that does not lead directly to a meaningful value without further investigation. These forms, such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), arise when evaluating limits where the numerator and denominator both approach zero or both approach infinity. Recognizing an indeterminate form is crucial because it signals the place where advanced techniques, like l'Hôpital's Rule, are needed to find a meaningful limit.

Consider the exercise where we find the limit \( \lim_{x \rightarrow 1} \frac{\ln x^{2}}{x^{2}-1} \). By substituting \( x = 1 \), the numerator becomes \( \ln(1^2) = 0 \) and the denominator \( 1^2 - 1 = 0 \). This results in the indeterminate form \( \frac{0}{0} \), which confirms that direct calculation won't work, and a more refined approach must be employed.

Identifying such forms ensures that we only apply l'Hôpital's Rule when necessary, preventing misuse and ensuring accurate solutions. After acknowledgment, moving forward with the correct calculus methods will yield the necessary result.

Limits

In mathematics, limits help us understand the behavior of a function as it approaches a particular point or as the variable approaches infinity. The concept is foundational in calculus and a crucial tool for defining derivatives and integrals.

When evaluating limits, several approaches can be used, such as direct substitution, factoring, or using properties of limits. However, when faced with indeterminate forms, strategies like l'Hôpital's Rule become essential. This rule allows us to differentiate the numerator and denominator separately, making it easier to solve otherwise complex limits.

In our exercise, after recognizing the indeterminate form, we apply l'Hôpital's Rule to evaluate \( \lim_{x \rightarrow 1} \frac{\ln x^{2}}{x^{2}-1} \). Differentiating the numerator and denominator, then simplifying, leads us to successfully compute the limit as \( x \to 1 \). Through this example, we see the power of limiting processes in solving seemingly difficult mathematical problems.

Calculus

Calculus is the branch of mathematics that studies change. It consists of two main branches: differential calculus, which involves the concept of derivatives, and integral calculus, which focuses on integrals. These tools are used to solve problems involving rates of change and areas under curves, among others.

In our exercise, calculus is the key to understanding and solving the limit problem. By utilizing techniques from differential calculus, specifically differentiation, l'Hôpital's Rule helps us break down complex indeterminate forms. The techniques in calculus allow us to solve problems that simple algebra cannot tackle alone.

Calculus equips us with methods to examine limits and other intriguing mathematical forms. It pushes the boundaries of mathematical investigations by considering infinite processes, providing solutions and insights beyond conventional static evaluations. From analyzing motions in physics to optimizing solutions in economics, calculus offers the tools to tackle a wide array of real-world problems.