To evaluate the limit

\[
\lim_{h \to 0} \frac{\sin(x + h) - \sin(x)}{h},
\]

we can apply a standard trigonometric identity and the definition of a derivative.

### Step-by-step process:

1. **Apply the sine addition formula**:
   \[
   \sin(x + h) = \sin(x)\cos(h) + \cos(x)\sin(h).
   \]
   Substituting this into the limit expression:
   \[
   \lim_{h \to 0} \frac{[\sin(x)\cos(h) + \cos(x)\sin(h)] - \sin(x)}{h}.
   \]

2. **Simplify the numerator**:
   \[
   = \lim_{h \to 0} \frac{\sin(x)(\cos(h) - 1) + \cos(x)\sin(h)}{h}.
   \]

3. **Split the limit into two parts**:
   \[
   = \lim_{h \to 0} \left[\sin(x) \cdot \frac{\cos(h) - 1}{h} + \cos(x) \cdot \frac{\sin(h)}{h}\right].
   \]

4. **Evaluate the limits** using standard results:
   - \(\lim_{h \to 0} \frac{\sin(h)}{h} = 1\),
   - \(\lim_{h \to 0} \frac{\cos(h) - 1}{h} = 0\).

   Substituting these results, we get:
   \[
   \lim_{h \to 0} \left[\sin(x) \cdot 0 + \cos(x) \cdot 1\right] = \cos(x).
   \]

### Final result:
\[
\lim_{h \to 0} \frac{\sin(x + h) - \sin(x)}{h} = \cos(x).
\]

This is the derivative of \(\sin(x)\), confirming the limit represents \(\frac{d}{dx}[\sin(x)] = \cos(x)\).

---

Would you like more details on any of the steps? 

Here are 5 related questions to explore further:

1. What is the limit of \((\cos(x+h)-\cos(x))/h\) as \(h \to 0\)?
2. How can we apply the definition of derivative to find the derivative of \(\cos(x)\)?
3. What is the derivative of \(\tan(x)\) using the limit definition?
4. How does the sine addition formula help in solving trigonometric limits?
5. How do trigonometric limits differ from polynomial limits?

**Tip**: Understanding the sine addition formula is crucial for solving complex trigonometric limits and derivatives.

Explore the solution for lim h→0 (sin(x+h) - sin(x))/h using trigonometric identities and derivatives. Understand the step-by-step process with the sine addition formula and learn how limits and derivatives apply to trigonometric functions. Ideal for undergraduate calculus students.

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