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To solve this problem, we will use the **limit definition of the derivative** to find the slope of the tangent line to the curve \( y = \sqrt{x} + 3 \) at the point \( (1, 2) \).

### Limit Definition of Derivative:
The limit definition of the derivative is given by:
\[
f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}
\]
Where \( f(x) = \sqrt{x} + 3 \).

### Step-by-Step Solution:

1. **Function at \( x = 1 \):** 
   \[
   f(1) = \sqrt{1} + 3 = 1 + 3 = 4
   \]

2. **Function at \( x + h \):**
   \[
   f(1 + h) = \sqrt{1 + h} + 3
   \]

3. **Substitute into the limit definition:**
   \[
   f'(1) = \lim_{h \to 0} \frac{f(1+h) - f(1)}{h} = \lim_{h \to 0} \frac{\left( \sqrt{1+h} + 3 \right) - 4}{h}
   \]
   Simplifying the numerator:
   \[
   f'(1) = \lim_{h \to 0} \frac{\sqrt{1+h} + 3 - 4}{h} = \lim_{h \to 0} \frac{\sqrt{1+h} - 1}{h}
   \]

4. **Rationalize the numerator:**
   Multiply both the numerator and the denominator by the conjugate \( \sqrt{1+h} + 1 \):
   \[
   f'(1) = \lim_{h \to 0} \frac{\left( \sqrt{1+h} - 1 \right) \cdot \left( \sqrt{1+h} + 1 \right)}{h \cdot \left( \sqrt{1+h} + 1 \right)}
   \]
   Using the difference of squares, \( (a - b)(a + b) = a^2 - b^2 \), we get:
   \[
   f'(1) = \lim_{h \to 0} \frac{(1 + h) - 1^2}{h \cdot (\sqrt{1+h} + 1)} = \lim_{h \to 0} \frac{h}{h \cdot (\sqrt{1+h} + 1)}
   \]

5. **Simplify:**
   \[
   f'(1) = \lim_{h \to 0} \frac{1}{\sqrt{1+h} + 1}
   \]

6. **Take the limit as \( h \to 0 \):**
   As \( h \to 0 \), \( \sqrt{1+h} \to 1 \), so:
   \[
   f'(1) = \frac{1}{1 + 1} = \frac{1}{2}
   \]

### Conclusion:
The slope of the tangent line to the curve \( y = \sqrt{x} + 3 \) at the point \( (1, 2) \) is \( \frac{1}{2} \).

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Would you like more details on any specific part of this solution? 

Here are 5 related questions:
1. What is the geometric interpretation of the derivative at a point?
2. How can you verify the slope of a tangent using the chain rule instead of the limit definition?
3. How does the process change if the function has a higher degree or is more complex?
4. Why is it necessary to rationalize the numerator when using the limit definition of the derivative?
5. How does the derivative of \( \sqrt{x} + 3 \) change as you move away from \( x = 1 \)?

**Tip:** Rationalizing the numerator is a common technique when dealing with square roots in limits.

Use the limit definition of the derivative to find the slope of the tangent line to the curve y = sqrt(x) + 3 at the point (1,2). Do not use differentiation formulas.

This problem involves finding the slope of the tangent line to the curve y = sqrt(x) + 3 at the point (1,2) using the limit definition of the derivative. Learn step-by-step how to rationalize the numerator and apply the limit to find the slope. Suitable for high school or college-level calculus students.

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