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To solve this problem, we need to find the derivative of \( f(x) = \sqrt{2x + 3} \) at \( x = 3 \), using the definition of the derivative:

\[
f'(3) = \lim_{h \to 0} \frac{f(3 + h) - f(3)}{h}
\]

Let's first calculate \( f(3) \):

\[
f(3) = \sqrt{2 \cdot 3 + 3} = \sqrt{6 + 3} = \sqrt{9} = 3
\]

Now, substitute \( f(3) = 3 \) and \( f(3 + h) = \sqrt{2(3 + h) + 3} = \sqrt{6 + 2h + 3} = \sqrt{9 + 2h} \) into the derivative formula:

\[
f'(3) = \lim_{h \to 0} \frac{\sqrt{9 + 2h} - 3}{h}
\]

Now, let's match this expression with the given options:

1. \(\lim_{h \to 0} \frac{\sqrt{2 \cdot 3 + 3 + h} - 3}{h} = \lim_{h \to 0} \frac{\sqrt{9 + h} - 3}{h}\)
2. \(\lim_{h \to 0} \frac{\sqrt{9 + h + 3} - 3}{h} = \lim_{h \to 0} \frac{\sqrt{12 + h} - 3}{h}\)
3. \(\lim_{h \to 0} \frac{\sqrt{9 + 2h} - 3}{h}\)
4. \(\lim_{h \to 0} \frac{\sqrt{9 + h} - 3}{h}\)

The correct option is **Option 3**: \(\lim_{h \to 0} \frac{\sqrt{9 + 2h} - 3}{h}\), as it matches our expression for \( f'(3) \). 

Would you like more details or have any questions?

### Related Questions
1. How do you find the derivative of a function using the limit definition?
2. What is the derivative of \( f(x) = \sqrt{2x + 3} \) for any \( x \)?
3. Why does rationalizing the numerator help in finding limits involving square roots?
4. Can you explain how to solve similar problems with different functions?
5. What is the importance of understanding derivatives in calculus?

**Tip:** When dealing with limits that involve square roots, try rationalizing the numerator to simplify the expression.

Si f(x) = √(2x + 3) entonces f'(3) es igual al siguiente límite.

This problem explores how to calculate the derivative of the function f(x) = √(2x + 3) at x = 3 using the limit definition of a derivative. The solution walks through the necessary steps to match the correct limit expression. Suitable for students in advanced high school or early college calculus.

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