https://file.aiquickdraw.com/mathbot/file/17267955592611crareuj.jpg

The function \( h(x) \) is given as:

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

We are asked to find the value of \( a \) such that \( \lim_{x \to 0} h(x) \) exists. 

To solve this, we need to check the limits of both pieces of the function as \( x \to 0 \) from the left (\( x \leq 0 \)) and from the right (\( x > 0 \)).

### 1. Right-hand limit (\( x \to 0^+ \)):
For \( x > 0 \), the function is:

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

As \( x \to 0^+ \), the first term \( x^2 \to 0 \). For the second term, apply L'Hopital's Rule, as it is of the indeterminate form \( \frac{0}{0} \) when \( x \to 0 \):

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

Differentiate the numerator and denominator with respect to \( x \):

- Numerator: \( \frac{d}{dx}(1 - \cos x) = \sin x \)
- Denominator: \( \frac{d}{dx}(2x \sin 2x) = 2\sin 2x + 4x \cos 2x \)

Thus, applying L'Hopital's Rule:

\[
\lim_{x \to 0^+} \frac{\sin x}{2\sin 2x + 4x \cos 2x} = \frac{0}{0 + 0} = 0
\]

So, the right-hand limit is:

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

### 2. Left-hand limit (\( x \to 0^- \)):
For \( x \leq 0 \), the function is:

\[
h(x) = \frac{a}{x - 2}
\]

As \( x \to 0^- \), the left-hand limit becomes:

\[
\lim_{x \to 0^-} \frac{a}{x - 2} = \frac{a}{-2}
\]

### 3. Equating limits:
For the limit to exist at \( x = 0 \), the right-hand limit must equal the left-hand limit. Therefore, we equate the two:

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

Solving for \( a \):

\[
a = 0
\]

Thus, the value of \( a \) such that \( \lim_{x \to 0} h(x) \) exists is \( \boxed{0} \).

Would you like more details or explanations on any part of the solution?

Here are 5 related questions for practice:
1. What is L'Hopital's Rule, and when can it be applied?
2. How do you calculate the limit of trigonometric functions as \( x \to 0 \)?
3. What are the conditions for the existence of a limit at a point?
4. Can the limit of a piecewise function be discontinuous at a point?
5. How do you solve indeterminate forms of type \( \frac{0}{0} \) using derivatives?

**Tip:** When solving limits involving trigonometric functions, always check for indeterminate forms and consider using L'Hopital’s Rule if applicable.

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

Solve a piecewise function involving limits and trigonometric identities. Use L'Hopital's Rule to determine the value of a that makes the limit of h(x) as x approaches 0 exist.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation