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The problem provided asks to determine the value of \( a \) so that the limit \(\lim_{x \to 0} h(x)\) exists for the given piecewise function:

\[
h(x) =
\begin{cases} 
x^2 + \frac{1 - \cos x}{2x \sin 2x} & x > 0 \\
\frac{a}{x - 2} & x \leq 0 
\end{cases}
\]

### Step 1: Evaluate the limit from the right-hand side (\(x \to 0^+\)).

For \(x > 0\), we have:

\[
h(x) = x^2 + \frac{1 - \cos x}{2x \sin 2x}
\]

- As \(x \to 0^+\), \(x^2 \to 0\).
- We need to analyze the second term: \(\frac{1 - \cos x}{2x \sin 2x}\).

Using standard limits:
- \(\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}\),
- \(\lim_{x \to 0} \frac{\sin 2x}{2x} = 1\).

Thus, we can evaluate the second term:

\[
\lim_{x \to 0} \frac{1 - \cos x}{2x \sin 2x} = \lim_{x \to 0} \frac{\frac{1 - \cos x}{x^2}}{\frac{2x \sin 2x}{x^2}} = 0
\]

Therefore, the right-hand limit becomes:

\[
\lim_{x \to 0^+} h(x) = 0 + 0 = 0.
\]

### Step 2: Evaluate the limit from the left-hand side (\(x \to 0^-\)).

For \(x \leq 0\), the function is \(h(x) = \frac{a}{x - 2}\).

As \(x \to 0^-\), we substitute directly into the expression:

\[
\lim_{x \to 0^-} h(x) = \frac{a}{0 - 2} = \frac{a}{-2} = -\frac{a}{2}.
\]

### Step 3: Set the limits equal to ensure continuity.

For the limit \(\lim_{x \to 0} h(x)\) to exist, the left-hand and right-hand limits must be equal:

\[
0 = -\frac{a}{2}.
\]

Solving for \(a\):

\[
a = 0.
\]

Thus, the value of \(a\) that ensures the limit exists is \(a = 0\).

---

Would you like further details or explanations?

Here are five related questions for practice:
1. What is the general condition for a piecewise function to have a limit at a point?
2. How does L'Hôpital's rule help in solving limits involving indeterminate forms?
3. Why does \(\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}\)?
4. How do you evaluate the limit of trigonometric functions as \(x \to 0\)?
5. Can a function be continuous but not differentiable at a point?

**Tip:** When solving piecewise functions involving limits, always check both sides independently before equating them.

Diberikan fungsi h, dengan h(x) = { x^2 + (1 - cos x) / (2x sin 2x) untuk x > 0, a / (x - 2) untuk x <= 0. Tentukan nilai a agar lim (x -> 0) h(x) ada.

Learn how to determine the value of 'a' that ensures the limit exists for a piecewise function involving x^2 and trigonometric limits. This problem covers concepts such as limits, continuity, and trigonometric limits, suitable for high school students in Grades 11-12.

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