Find limit as x approaches 0, 1 - cos^2 (x) / 4x

We are tasked with finding the limit:

\[
\lim_{{x \to 0}} \frac{1 - \cos^2(x)}{4x}
\]

### Step 1: Simplify the expression
We can simplify \(1 - \cos^2(x)\) using the identity:

\[
1 - \cos^2(x) = \sin^2(x)
\]

Thus, the expression becomes:

\[
\lim_{{x \to 0}} \frac{\sin^2(x)}{4x}
\]

### Step 2: Split the fraction
Now, rewrite the fraction as:

\[
\lim_{{x \to 0}} \frac{\sin^2(x)}{4x} = \lim_{{x \to 0}} \frac{\sin(x)}{x} \cdot \frac{\sin(x)}{4}
\]

### Step 3: Apply known limits
We know that:

\[
\lim_{{x \to 0}} \frac{\sin(x)}{x} = 1
\]

Therefore, the limit becomes:

\[
\lim_{{x \to 0}} \frac{\sin(x)}{x} \cdot \frac{\sin(x)}{4} = 1 \cdot \frac{0}{4} = 0
\]

### Final Answer:
\[
\lim_{{x \to 0}} \frac{1 - \cos^2(x)}{4x} = 0
\]

Would you like further details on any step?

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Here are five related questions to explore further:
1. What is the limit of \(\frac{\sin(x)}{x}\) as \(x \to 0\), and why?
2. How does L'Hopital's rule apply to indeterminate forms like \(\frac{0}{0}\)?
3. Can you find the limit of \(\frac{1 - \cos(x)}{x^2}\) as \(x \to 0\)?
4. How can Taylor series expansions help in solving limits?
5. What are other common trigonometric limits you should know?

**Tip:** When solving limits involving trigonometric functions, use known trigonometric limits and identities to simplify complex expressions.

Learn how to find the limit of (1 - cos^2(x)) / 4x as x approaches 0. This problem uses key trigonometric identities like sin^2(x) = 1 - cos^2(x) and involves limit theorems such as lim(x -> 0) (sin(x) / x) = 1. Suitable for students in Grades 11-12.

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