To find the derivative of $f(x) = x + \sqrt{x}$ using the limit definition, we start by using the formula:

$f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}$

Step 1: Substitute $f(x) = x + \sqrt{x}$ into $f(a + h)$ and $f(a)$

1. Calculate $f(a + h)$: $f(a + h) = (a + h) + \sqrt{a + h}$

2. Calculate $f(a)$: $f(a) = a + \sqrt{a}$

Step 2: Substitute into the Limit Definition

$f'(a) = \lim_{h \to 0} \frac{(a + h + \sqrt{a + h}) - (a + \sqrt{a})}{h}$

This simplifies to:

$f'(a) = \lim_{h \to 0} \frac{h + \sqrt{a + h} - \sqrt{a}}{h}$

Step 3: Separate Terms

Split the limit into two parts:

$f'(a) = \lim_{h \to 0} \left( \frac{h}{h} + \frac{\sqrt{a + h} - \sqrt{a}}{h} \right)$

1. The first term, $\frac{h}{h}$, simplifies to 1.

2. The second term requires us to rationalize the numerator:

   $\frac{\sqrt{a + h} - \sqrt{a}}{h} \times \frac{\sqrt{a + h} + \sqrt{a}}{\sqrt{a + h} + \sqrt{a}} = \frac{(a + h) - a}{h(\sqrt{a + h} + \sqrt{a})}$

   Simplify the numerator:

   $= \frac{h}{h(\sqrt{a + h} + \sqrt{a})} = \frac{1}{\sqrt{a + h} + \sqrt{a}}$

Step 4: Take the Limit

Now, substitute $h = 0$ in the second term:

$\lim_{h \to 0} \frac{1}{\sqrt{a + h} + \sqrt{a}} = \frac{1}{2\sqrt{a}}$

Final Answer

Combining the terms, we get:

$f'(a) = 1 + \frac{1}{2\sqrt{a}}$

Thus, the derivative of $f(x) = x + \sqrt{x}$ at any point $a \in [0, \infty)$ is:

$f'(a) = 1 + \frac{1}{2\sqrt{a}}$

Would you like a detailed explanation of any part?

---

1. What is the geometric interpretation of the derivative found here?
2. How does the derivative behave as $a$ approaches infinity?
3. Why is rationalizing the numerator necessary in limit problems involving square roots?
4. Can this derivative be obtained using rules of differentiation instead of the limit definition?
5. How does the derivative change for functions similar to $f(x) = x + \sqrt{x}$?

Tip: When working with square roots in limits, rationalization often simplifies the problem by eliminating the radical in the numerator.