The problem involves the following limit:

$\lim_{x \to a} \frac{\sin^2(x) - \sin^2(a)}{\ln(x) - \ln(a)}.$

Step 1: Simplify the numerator

Using the identity $\sin^2(x) - \sin^2(a) = (\sin(x) - \sin(a))(\sin(x) + \sin(a))$, we rewrite the numerator: $\sin^2(x) - \sin^2(a) = (\sin(x) - \sin(a))(\sin(x) + \sin(a)).$

Step 2: Apply L'Hôpital's Rule

Both the numerator and denominator approach $0$ as $x \to a$, so we can apply L'Hôpital's Rule. First, differentiate the numerator and denominator:

• Derivative of the numerator: $\frac{d}{dx}\left[\sin^2(x) - \sin^2(a)\right] = 2\sin(x)\cos(x).$

• Derivative of the denominator: $\frac{d}{dx}\left[\ln(x) - \ln(a)\right] = \frac{1}{x}.$

The limit becomes: $\lim_{x \to a} \frac{2\sin(x)\cos(x)}{\frac{1}{x}} = \lim_{x \to a} 2x\sin(x)\cos(x).$

Step 3: Simplify further

Using the double-angle identity $\sin(2x) = 2\sin(x)\cos(x)$, we rewrite: $2x\sin(x)\cos(x) = x\sin(2x).$

As $x \to a$, substitute $x = a$: $\lim_{x \to a} x\sin(2x) = a\sin(2a).$

Final Answer:

$\lim_{x \to a} \frac{\sin^2(x) - \sin^2(a)}{\ln(x) - \ln(a)} = a\sin(2a).$

Would you like further clarification or additional examples?

Related Questions:

1. What is L'Hôpital's Rule and when is it applied?
2. How can trigonometric identities simplify limits?
3. What happens if $a = 0$ in this limit?
4. Can this problem be solved without using L'Hôpital's Rule?
5. How does the chain rule apply in derivative computations here?

Tip:

Always verify whether the numerator and denominator both approach $0$ or $\pm\infty$ before applying L'Hôpital's Rule.