Problem 5 - University Calculus: Early Transcendentals

Question

Use I'Hópital's Rule to evaluate the limit. Then evaluate the limit using a method studied in Chapter 2. $$\lim _{x \rightarrow 0} \frac{1-\cos x}{x^{2}}$$

Step-by-Step Solution

Answer

The limit is \( \frac{1}{2} \).

Step 1: Identify the Indeterminate Form

The expression \( \lim_{x \rightarrow 0} \frac{1-\cos x}{x^2} \) seems to be an indeterminate form \( \frac{0}{0} \). As \( x \to 0 \), \( 1 - \cos x \to 0 \) and \( x^2 \to 0 \). This justifies the use of L'Hopital's Rule.

Step 2: Apply L'Hopital's Rule

L'Hopital's Rule states that if the limit \( \lim_{x \to a} \frac{f(x)}{g(x)} \) results in an indeterminate form \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), then it is equal to \( \lim_{x \to a} \frac{f'(x)}{g'(x)} \), assuming this limit exists. Differentiate the numerator and the denominator: \( f(x) = 1 - \cos x \) and \( g(x) = x^2 \).- Derivative of the numerator, \( f'(x) = \sin x \).- Derivative of the denominator, \( g'(x) = 2x \).

Step 3: Evaluate the New Limit

Now evaluate \( \lim_{x \to 0} \frac{\sin x}{2x} \). Substituting into L'Hopital's Rule gives:\[\lim_{x \to 0} \frac{\sin x}{2x} = \lim_{x \to 0} \frac{1}{2} \cdot \frac{\sin x}{x} = \frac{1}{2} \cdot 1 = \frac{1}{2}\]This follows from the standard limit \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \).

Step 4: Verify Using Series Expansion Method

Using the Maclaurin series expansion, \( \cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \cdots \), substitute into the original limit:\[1 - \cos x = x^2 \left(\frac{1}{2} - \frac{x^2}{24} + \cdots \right)\]Thus,\[\lim_{x \to 0} \frac{1-\cos x}{x^{2}} = \lim_{x \to 0} \left( \frac{x^2 \left(\frac{1}{2} - \frac{x^2}{24} + \cdots \right)}{x^2}\right) = \frac{1}{2}\]

Step 5: Conclusion

Both methods, L'Hopital's Rule and series expansion, provide the same result. Hence, \( \lim_{x \to 0} \frac{1-\cos x}{x^{2}} = \frac{1}{2} \).

Key Concepts

Indeterminate Forms

In calculus, an indeterminate form is a mathematical expression that takes on different values depending on the circumstances, particularly at limits. When we say the limit results in an indeterminate form, it means we can't directly determine the limit without further evaluation. A common indeterminate form is \( \frac{0}{0} \).
When you substitute the point where the limit occurs into the functions in both the numerator and denominator, and you get zero for both, this indicates an indeterminate form like \( \frac{0}{0} \).
For example, with the limit \( \lim_{x \to 0} \frac{1-\cos x}{x^2} \), both \( 1 - \cos x \) and \( x^2 \) approach zero as \( x \) approaches zero. This causes the expression to take an indeterminate form. Recognizing these forms is crucial as it dictates the methods you can use to solve the problem, like L'Hôpital's Rule or series expansion.

Limit Evaluation

Limit evaluation is the process of determining what a function approaches as we get near a certain point. This involves determining the behavior of the function close to a particular \( x \)-value, often to find a value the function seems to trend towards. There are various ways to evaluate limits:

• L'Hôpital's Rule: Useful for indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). By taking the derivatives of the numerator and denominator, it allows a new limit evaluation under specific conditions.
• Simplification Techniques: Manipulating the expression to simplify it into a more straightforward form that can be easily evaluated, such as factoring or expanding.
• Standard Limit Results: Using known limit results like \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \) to evaluate more complex limits.

By evaluating \( \lim_{x \to 0} \frac{1-\cos x}{x^2} \) using L'Hôpital's Rule, we find \( \lim_{x \to 0} \frac{\sin x}{2x} = \frac{1}{2} \). This includes understanding how standard limit results apply in this context, such as how \( \frac{\sin x}{x} \to 1 \) as \( x \to 0 \).

Maclaurin Series

The Maclaurin series is a type of Taylor series expanded about \( x = 0 \). It provides an infinite sum of terms that approximates functions near this point. For example, the series for \( \cos x \) is:\[ \cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \cdots \]This expansion is very helpful in limit evaluations, especially when dealing with indeterminate forms. By substituting the series expansion into the limit \( \lim_{x \to 0} \frac{1-\cos x}{x^2} \), the series simplifies the process, allowing terms to cancel and reduce complexity.
Once expanded, \( 1 - \cos x \) transforms into \( x^2 \left( \frac{1}{2} - \frac{x^2}{24} + \cdots \right) \).
When dividing by \( x^2 \), simplification leads directly to a result of \( \frac{1}{2} \), the limit's value. Understanding and applying series like the Maclaurin series help overcome complicated evaluations in calculus, especially around \( x = 0 \), offering a powerful tool for approximations.