Problem 22 - Single Variable Calculus

Question

evaluate the limit using l'Hôpital's Rule if appropriate. $$ \lim _{x \rightarrow 0} \frac{\ln \left(x^{2}+1\right)}{\cos x-1} $$

Step-by-Step Solution

Answer

The limit as x approaches 0 of \(\frac{\ln (x^2 + 1)}{\cos x - 1}\) is 2, and we used l'Hôpital's Rule twice to evaluate the limit.

Step 1: Determine the initial limit

We need to find the limits of the numerator and denominator when x approaches 0. For the numerator: $$ \lim_{x \rightarrow 0} \ln(x^2 + 1) $$ For the denominator: $$ \lim_{x \rightarrow 0} (\cos x - 1) $$ Let's compute the limits: For the numerator: $$ \lim_{x \rightarrow 0} \ln(x^2 + 1) = \ln(0^2 + 1) = \ln(1) = 0 $$ For the denominator: $$ \lim_{x \rightarrow 0} (\cos x - 1) = \cos(0) - 1 = 1 - 1 = 0 $$ The limit has an indeterminate form of \(\frac{0}{0}\). Thus, we can apply l'Hôpital's Rule.

Step 2: Apply l'Hôpital's Rule

According to l'Hôpital's Rule, if the limit is in the form \(\frac{0}{0}\), we can take the derivative of the numerator and denominator, and then find the limit: $$ \lim_{x \rightarrow 0} \frac{\ln (x^2 + 1)}{\cos x - 1} = \lim_{x \rightarrow 0} \frac{\frac{d}{dx} (\ln(x^2 + 1))}{\frac{d}{dx} (\cos x - 1)} $$ Let's first find the derivatives of the numerator and the denominator separately. For the numerator, using the chain rule: $$ \frac{d}{dx} (\ln(x^2 + 1)) = \frac{1}{x^2 + 1} \cdot 2x = \frac{2x}{x^2 + 1} $$ For the denominator: $$ \frac{d}{dx} (\cos x - 1) = -\sin x $$ Now, let's substitute the derivatives back into the limit: $$ \lim_{x \rightarrow 0} \frac{\frac{2x}{x^2 + 1}}{-\sin x} $$

Step 3: Evaluate the new limit

Let's compute the limit of the new expression as x approaches 0: $$ \lim_{x \rightarrow 0} \frac{\frac{2x}{x^2 + 1}}{-\sin x} = \frac{\frac{2(0)}{0^2 + 1}}{-\sin (0)} = \frac{0}{0} $$ The new limit is still in the indeterminate form of \(\frac{0}{0}\), so we will need to apply l'Hôpital's Rule again.

Step 4: Apply l'Hôpital's Rule again

Let's take the derivatives of the new numerator and denominator and find the limit: $$ \lim_{x \rightarrow 0} \frac{\frac{2x}{x^2+1}}{-\sin x} = \lim_{x \rightarrow 0} \frac{\frac{d}{dx} (\frac{2x}{x^2 + 1})}{\frac{d}{dx} (-\sin x)} $$ Finding the derivatives: For the numerator, using the quotient rule: $$ \frac{d}{dx} (\frac{2x}{x^2 + 1}) = \frac{(x^2 + 1) \cdot 2 - 2x \cdot 2x}{(x^2 + 1)^2} = \frac{2x^2 + 2 - 4x^2}{(x^2 + 1)^2} = \frac{-2x^2 + 2}{(x^2 + 1)^2} $$ For the denominator: $$ \frac{d}{dx} (-\sin x) = -\cos x $$ Now, let's substitute the derivatives back into the limit: $$ \lim_{x \rightarrow 0} \frac{\frac{-2x^2 + 2}{(x^2 + 1)^2}}{-\cos x} $$

Step 5: Evaluate the final limit

Let's compute the limit of the final expression as x approaches 0: $$ \lim_{x \rightarrow 0} \frac{\frac{-2x^2 + 2}{(x^2 + 1)^2}}{-\cos x} = \frac{\frac{-2(0)^2 + 2}{(0^2 + 1)^2}}{-\cos (0)} = \frac{2}{1} = 2 $$ The limit as x approaches 0 of \(\frac{\ln (x^2 + 1)}{\cos x - 1}\) is 2, and we used l'Hôpital's Rule to find the solution.

Key Concepts

Limits and Continuity

Understanding limits is a fundamental part of calculus, especially when exploring the behavior of functions as inputs approach a certain value.

In calculus education, the concept of a limit is introduced as a way to find the value that a function approaches as the variable gets infinitely close to a point. A key application of limits is in determining the continuity of a function at a point. A function is said to be continuous at a point if the limit as x approaches the point is equal to the function's value at that point.

Enhancing the Understanding of Limits

To deepen your understanding, consider visualizing functions on a graph to observe how they behave near the point of interest. Additionally, practicing various types of limit problems, such as those with infinity, zero, and indeterminate forms, can build a more robust conceptual foundation.

Indeterminate Forms

Indeterminate forms occur when evaluating a limit leads to an expression that is not immediately obvious in determining the limit’s value. Common indeterminate forms include \(\frac{0}{0}\), \(\frac{\infty}{\infty}\), \(0 \cdot \infty\), \(\infty - \infty\), \(0^0\), \(\infty^0\), and \(1^\infty\).

Addressing Indeterminate Forms

In cases such as \(\frac{\ln (x^2 + 1)}{\cos x - 1}\) approaching \(\frac{0}{0}\), l'Hôpital's Rule becomes particularly useful. This rule allows us to differentiate the numerator and the denominator independently before taking the limit again. This process may need to be repeated until the limit can be determined or shown to not exist. By practicing exercises involving indeterminate forms, students can learn to recognize and confidently resolve these challenging limits.

Derivative Applications

Derivatives are not just about finding slopes or rates of change; they are instrumental in solving a range of problems in calculus. One notable application is l'Hôpital's Rule, which uses derivatives to resolve limits involving indeterminate forms.

When faced with a limit that evaluates to \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), we can apply the rule to differentiate the numerator and denominator. The derivative application requires a solid grasp of rules such as the chain rule, product rule, and quotient rule, as seen in the textbook solution provided.

Practical Tips on Derivatives

To better understand derivatives, start with simple functions and gradually work towards more complex examples. Use multiple representation forms, like graphs and tables, to illustrate how the derivative reflects changes in the original function.

Calculus Education

Calculus can be a challenging subject for many students, but with the right approach, it can be made accessible and enjoyable. Providing step-by-step solutions, as shown in the provided exercise, is one effective method of instruction.

However, to truly grasp the concepts, students should be encouraged to engage in active learning. This includes solving a variety of problems on their own, exploring real-world applications of calculus, and utilizing interactive tools to visualize concepts.

Strategies for Effective Learning

To enhance calculus education, introduce concepts progressively and ensure that students have a firm understanding of foundational topics, like algebra and trigonometry, before advancing. Encourage questions and discussions to facilitate a deeper understanding, applying calculus to solve practical problems whenever possible.