Problem 12 - Calculus Single Variable

Question

Use l'Hôpital's Rule to find the limit, if it exists. \(\lim _{x \rightarrow 0} \frac{\sin ^{2}(3 x)}{1-\cos (4 x)}\)

Step-by-Step Solution

Answer

The limit is \(\frac{9}{8}\).

Step 1: Form Check for Indeterminate Form

First, determine if the limit leads to an indeterminate form. We set \(x \rightarrow 0\) in the function \(f(x) = \frac{\sin^2(3x)}{1-\cos(4x)}\). At \(x=0\), \(\sin(3x) = 0\) and \(\cos(4x) = 1\). Therefore, the expression simplifies to \(\frac{0}{0}\), showing it is an indeterminate form suitable for l'Hôpital's Rule.

Step 2: Differentiate Numerator and Denominator

Apply l'Hôpital's Rule by differentiating both the numerator and the denominator separately. The derivative of the numerator is obtained using the chain and power rules: \(\frac{d}{dx}[\sin^2(3x)] = 2\sin(3x)\cdot\cos(3x)\cdot3 = 6\sin(3x)\cos(3x)\). The derivative of the denominator using the chain rule is \(\frac{d}{dx}[1-\cos(4x)] = 4\sin(4x)\).

Step 3: Apply l'Hôpital's Rule and Simplify

Replace the original fraction with the derivatives from Step 2: \[\lim _{x \rightarrow 0} \frac{6\sin(3x)\cos(3x)}{4\sin(4x)}\]. Simplify the expression. Since \(\sin(3x)\cos(3x) = \frac{1}{2} \sin(6x)\), rewrite the limit: \[\lim _{x \rightarrow 0} \frac{6 \cdot \frac{1}{2}\sin(6x)}{4\sin(4x)} = \lim _{x \rightarrow 0} \frac{3 \sin(6x)}{4 \sin(4x)}\].

Step 4: Evaluate the Simplified Limit

This limit can now be further simplified. Factor out common terms: \(\frac{3}{4}\cdot \lim_{x \rightarrow 0} \frac{\sin(6x)}{\sin(4x)}\). Rewrite this as: \(\frac{3}{4}\cdot \lim_{x \rightarrow 0} \frac{6x}{4x}\) because \(\lim_{x \rightarrow 0} \frac{\sin(kx)}{kx} = 1\). This simplifies to \(\frac{3}{4} \cdot \frac{6}{4} = \frac{18}{16}\), which reduces to \(\frac{9}{8}\).

Step 5: Confirm the Convergent Limit

Verify that substitutions or alternate considerations do not affect the simplified limit: Since each assumption (such as continuity and reorder) holds, and the result is consistent by another limit substitution, we are assured that our limit evaluation is accurate.

Key Concepts

Indeterminate Forms

When solving limits in calculus, we often encounter expressions that can't be directly evaluated, known as indeterminate forms. These forms occur when both the numerator and denominator approach 0 or infinity simultaneously. The classic example is the \(\frac{0}{0}\) form.
This happens frequently with trigonometric limits and expressions involving subtraction that tends to zero. If we find our limit expression evaluates to \(\frac{0}{0}\), we can't simply resolve it. Instead, we need to apply advanced methods like l'Hôpital's Rule to find the desired limit.
Recognizing an indeterminate form is the first step to choosing an appropriate technique to resolve it. It's the gateway to using l'Hôpital's Rule effectively.

Differentiation

Differentiation is the calculus technique of finding the derivative. It allows us to understand how a function changes at a given point, which is crucial when using l'Hôpital's Rule. In our example, to resolve the indeterminate form, we differentiate both the numerator and the denominator.

• The numerator \((\sin^2(3x))\) involves both power and chain rules. We set up the function in a way to apply these rules correctly: \(2\sin(3x)\cos(3x)\cdot3\), which simplifies to \(6\sin(3x)\cos(3x)\).
• The denominator \((1-\cos(4x))\) is simplified using the chain rule. The derivative is \(4\sin(4x)\).

Remember, the power rule deals with expressions like \(x^n\), and the chain rule helps with functions inside other functions, like in our trigonometric setup.

Trigonometric Limits

Trigonometric limits often appear tricky due to oscillating behaviors of trigonometric functions. Understanding their limits is essential when solving calculus problems that involve sine and cosine.
One often-used rule is \(\lim_{x \rightarrow 0} \frac{\sin(kx)}{kx} = 1\). This helps simplify expressions during limit evaluation. In our example, after applying l'Hôpital's Rule once, \(\frac{\sin(6x)}{\sin(4x)}\) is considered. Such expressions are rewritten using the known sine limit.
This limit rule is especially handy to resolve complex trigonometric expressions efficiently without extra computation. It's a fundamental tool in any calculus student's toolkit.

Calculus Problems

Calculus problems can often seem daunting. They require familiarity with a variety of concepts, like differentiation and limits. Trigonometric limits, particularly, can test even the seasoned learner.
Breaking problems down into manageable steps is key. Start by checking if there's an indeterminate form. If there is, consider if l'Hôpital's Rule is applicable. Evaluate derivatives carefully, looking to simplify at each stage.

• Identify core steps such as checking for indeterminate forms and simplifying expressions using known limits.
• Use differentiation rules accurately—chain, product, power—this impacts the solution directly.
• Practice regularly with different types of problems to reinforce understanding.

Each calculus problem is an opportunity to master these concepts. With patience and practice, solving complex problems becomes intuitive.