To find $f'(a)$ using the limit definition of the derivative, we start with:

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

where $f(x) = x + \sqrt{x}$.

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

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

3. Substitute into the limit definition: $f'(a) = \lim_{h \to 0} \frac{(a+h + \sqrt{a+h}) - (a + \sqrt{a})}{h}$

4. Simplify the expression: Distribute and cancel $a$ terms: $f'(a) = \lim_{h \to 0} \frac{h + \sqrt{a+h} - \sqrt{a}}{h}$ Separate the terms: $f'(a) = \lim_{h \to 0} \left( \frac{h}{h} + \frac{\sqrt{a+h} - \sqrt{a}}{h} \right)$ Simplify the first term: $f'(a) = \lim_{h \to 0} \left( 1 + \frac{\sqrt{a+h} - \sqrt{a}}{h} \right)$

5. Simplify the second term using conjugates: Multiply the numerator and denominator by the conjugate $\sqrt{a+h} + \sqrt{a}$: $f'(a) = \lim_{h \to 0} \left( 1 + \frac{(\sqrt{a+h} - \sqrt{a})(\sqrt{a+h} + \sqrt{a})}{h (\sqrt{a+h} + \sqrt{a})} \right)$ This simplifies to: $f'(a) = \lim_{h \to 0} \left( 1 + \frac{a + h - a}{h (\sqrt{a+h} + \sqrt{a})} \right)$ $= \lim_{h \to 0} \left( 1 + \frac{h}{h (\sqrt{a+h} + \sqrt{a})} \right)$ Cancel $h$ in the second term: $f'(a) = \lim_{h \to 0} \left( 1 + \frac{1}{\sqrt{a+h} + \sqrt{a}} \right)$ As $h \to 0$, $\sqrt{a+h} \to \sqrt{a}$, so: $f'(a) = 1 + \frac{1}{2\sqrt{a}}$

Final Answer:

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

Would you like a deeper breakdown of any step or further questions on this?

Here are some follow-up questions:

1. What is the significance of using the limit definition in calculus?
2. How would you find $f'(a)$ if $f(x)$ had a different form, like $f(x) = x^2 + \sqrt{x}$?
3. How does $f'(a)$ behave as $a \to 0$ and as $a \to \infty$?
4. How would the approach change if $f(x)$ included a negative square root term?
5. Can you compute $f'(a)$ at a specific point, such as $a = 4$?

Tip: When using the limit definition of a derivative, simplifying complex terms with conjugates can help remove indeterminate forms.