Problem 6 - Essential Calculus Early Transcendentals

Question

\(1-38=\) Find the limit. Use l'Hospital's Rule where appropriate. If there is a more elementary method, consider using it. If l'Hospital's Rule doesn't apply, explain why. $$\lim _{x \rightarrow 0} \frac{x^{2}}{1-\cos x}$$

Step-by-Step Solution

Answer

The limit is 2.

Step 1: Identify the Form of the Limit

The given limit is \( \lim_{x \to 0} \frac{x^2}{1 - \cos x} \). As \(x \to 0\), the numerator \(x^2 \to 0\) and the expression \(1 - \cos x \to 0\) because \(\cos 0 = 1\). So, the limit is in the indeterminate form \(\frac{0}{0}\), which suggests that l'Hôpital's Rule can be applied.

Step 2: Apply l'Hôpital's Rule

Since the limit is in the indeterminate form \(\frac{0}{0}\), we differentiate the numerator and the denominator separately. The derivative of \(x^2\) is \(2x\), and the derivative of \(1 - \cos x\) is \(\sin x\). Thus, the limit becomes: \[\lim_{x \to 0} \frac{2x}{\sin x}\]

Step 3: Simplify and Re-evaluate the Limit

Next, we simplify and re-evaluate the limit \( \lim_{x \to 0} \frac{2x}{\sin x} \). We know that \( \frac{\sin x}{x} \to 1 \) as \( x \to 0 \). Therefore, by rewriting the expression, we have:\[\lim_{x \to 0} 2 \cdot \frac{x}{\sin x} = 2 \cdot \lim_{x \to 0} \frac{x}{\sin x} = 2 \cdot \frac{1}{1} = 2\]

Step 4: Conclusion of the Limit Evaluation

Based on our calculation, the limit \( \lim_{x \to 0} \frac{x^2}{1 - \cos x} \) evaluates to 2, using l'Hôpital's Rule and the behavior of \( \frac{\sin x}{x}\) as \(x\) approaches 0.

Key Concepts

Indeterminate Form

In calculus, the concept of an **indeterminate form** is crucial when evaluating limits. These forms usually result from direct substitution in a limit expression that yields undefined or ambiguous values. Common indeterminate forms include \(\frac{0}{0}\), \(\frac{\infty}{\infty}\), \(0 \cdot \infty\), and others.

• The limit in this problem initially presents the form \(\frac{0}{0}\) as both the numerator \(x^2\) and the denominator \(1 - \cos x\) tend towards 0 as \(x\) approaches 0.
• This signals us to use other methods, like l'Hôpital's Rule, to evaluate the limit.

Understanding this form helps us to choose an appropriate mathematical tool for resolving ambiguity and successfully finding the limit.

Limit Evaluation

Evaluating limits involves determining the behavior of a function as the variable approaches a particular value. For this exercise, we aim to find the limit of \( \frac{x^2}{1 - \cos x} \) as \( x \to 0 \).When a direct substitution into a limit expression leads to an indeterminate form, other techniques such as:

• l'Hôpital's Rule
• Algebraic manipulation
• Substitution
• Trigonometric identities

can be employed to resolve the indeterminacy.Tackling this kind of problem requires identifying the form of the expression, which then dictates using rules like l'Hôpital's. This confirms whether a substitution or differentiation needs application to evaluate the limit without ambiguity.

Differentiation

Differentiation plays a crucial role in evaluating limits that yield indeterminate forms. In our problem, knowing how to differentiate the functions in the numerator and denominator is essential. To apply **l'Hôpital's Rule**, we differentiate the numerator \(x^2\) and the denominator \(1 - \cos x\):

• The derivative of \(x^2\) is \(2x\).
• The derivative of \(1 - \cos x\) is \(\sin x\).

By substituting these derivatives back into the limit expression, we transform the original tricky problem into a simpler one: \(\lim_{x \to 0} \frac{2x}{\sin x}\).differentiation is not solely about obtaining rates of change; it's a strategy to unravel complex limits by systematically simplifying expressions until we reach a workable answer.

Trigonometric Limits

Trigonometric functions often appear in calculus problems, and there are well-known limits associated with them. A key **trigonometric limit** to know is \( \lim_{x \to 0} \frac{\sin x}{x} = 1\). This limit is a cornerstone when dealing with limits involving sine and cosine, especially within l'Hôpital's framework. In our problem:

• We redefine \( \lim_{x \to 0} \frac{2x}{\sin x} \) using \(\frac{x}{\sin x} \to 1\).
• This simplifies to \( 2 \cdot 1 = 2 \), giving the final limit result.

Familiarity with trigonometric limits enhances our toolkit, allowing us to evaluate seemingly complex expressions with confidence and ease.