Problem 22 - Calculus Single Variable

Question

Apply l'Hôpital's Rule repeatedly (when needed) to evaluate the given limit, if it exists. \(\lim _{x \rightarrow 0} \frac{1-\cos (4 x)}{\sin ^{2}(x)}\)

Step-by-Step Solution

Answer

The limit is \(-8\).

Step 1: Confirm Indeterminate Form

First, check if the limit results in an indeterminate form. Substitute \(x = 0\) into the expression \(\frac{1-\cos(4x)}{\sin^2(x)}\). Both the numerator \(1 - \cos(4x)\) approaches \(0\) and the denominator \(\sin^2(x)\) approaches \(0\) as \(x\) approaches zero. Hence, \(\frac{0}{0}\) confirms an indeterminate form suitable for l'Hôpital's Rule.

Step 2: Apply l'Hôpital's Rule

Apply l'Hôpital's Rule, which states that if the limit \(\lim_{x \to c} \frac{f(x)}{g(x)}\) results in \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), then it equals \(\lim_{x \to c} \frac{f'(x)}{g'(x)}\) if this limit exists.Differentiate the numerator: \(-\sin(4x) \times 4 = -4\sin(4x)\).Differentiate the denominator: \(2\sin(x)\cos(x) = \sin(2x)\).The new limit to evaluate is \(\lim_{x \to 0} \frac{-4\sin(4x)}{\sin(2x)}\).

Step 3: Evaluate Simplified Limit

The limit \(\lim_{x \to 0} \frac{-4\sin(4x)}{\sin(2x)}\) can still present an indeterminate form since both \(\sin(4x)\) and \(\sin(2x)\) approach zero, resulting in another \(\frac{0}{0}\) situation.The functions can simplify, or we can apply l'Hôpital's Rule again, differentiating:Differentiate the numerator: \(-4 \times 4\cos(4x) = -16\cos(4x)\).Differentiate the denominator: \(2\cos(2x)\).The limit now becomes \(\lim_{x \to 0} \frac{-16\cos(4x)}{2\cos(2x)}\).

Step 4: Simplify and Calculate Final Limit

Substitute \(x = 0\) into the expression \(\frac{-16\cos(4x)}{2\cos(2x)}\):\(\cos(4 \times 0) = 1\) and \(\cos(2 \times 0) = 1\), so the expression simplifies to \(\frac{-16 \times 1}{2 \times 1} = \frac{-16}{2} = -8\).The limit evaluates to \(-8\).

Key Concepts

Indeterminate Forms

When evaluating limits, particularly in calculus, you might encounter expressions that don't give a clear result at first glance. Indeterminate forms arise when substituting a value into a function results in forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). These forms indicate that more analysis is needed to find the actual limit because the direct substitution does not work. In our exercise, substituting \( x = 0 \) resulted in \( \frac{0}{0} \), since \( 1 - \cos(4x) \) and \( \sin^2(x) \) both approach zero as \( x \) approaches zero. Recognizing indeterminate forms is crucial, as it tells us that we need a strategy, like l'Hôpital's Rule, to resolve the limit correctly.

Derivatives

Derivatives represent the rate of change of a function. In the context of l'Hôpital's Rule, derivatives help us simplify and evaluate limits of indeterminate forms. The rule relies on taking the derivative of both the numerator and the denominator once you spot a \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) situation. For instance:

• The derivative of \( 1-\cos(4x) \) with respect to \( x \) is \( -4\sin(4x) \).
• The derivative of \( \sin^2(x) \) converts to \( \sin(2x) \).

After calculating these derivatives, you can attempt the limit again. If it's still indeterminate, you may need to apply the rule multiple times.

Limit Evaluation

Limit evaluation is a fundamental concept in calculus, used to understand the behavior of functions as they approach specific points. It's about finding the output value a function approaches as the input gets closer to a particular number. When we use techniques such as l'Hôpital's Rule, we are often simplifying functions to evaluate their limits without directly substituting the problematic point. In our exercise, we repeatedly differentiated the expressions until reaching a straightforward form. Finally, substituting back into the simplified expression, \[ \lim_{x \to 0} \frac{-16\cos(4x)}{2\cos(2x)} \]showed a clear value, as \( \cos(0) = 1 \), leading to the conclusion of \(-8\).

Calculus Techniques

Calculus is full of techniques that allow us to solve complex problems, like limit evaluations. l'Hôpital's Rule is one of these essential methods. It's specifically designed to deal with indeterminate forms by letting us take derivatives instead of substituting directly. This approach can untangle the complexities of expressions that initially seem unsolvable. Using l'Hôpital's Rule efficiently requires practice:

• First, identify the indeterminate form.
• Apply derivatives to numerator and denominator.
• Recheck the form and reapply the rule if necessary.
• Simplify step by step until a determinable form is reached.

This stepwise approach is part of broader calculus techniques, helping solve not just for limits but for understanding functions deeply.