To find $g'(x)$ using the limit definition of a derivative for the function $g(x) = \sqrt{x} - 2$, we can follow these steps:

The limit definition of a derivative is:

$g'(x) = \lim_{{h \to 0}} \frac{g(x+h) - g(x)}{h}$

Step 1: Substitute the function $g(x) = \sqrt{x} - 2$ into the limit definition.

$g'(x) = \lim_{{h \to 0}} \frac{\left( \sqrt{x+h} - 2 \right) - \left( \sqrt{x} - 2 \right)}{h}$

Step 2: Simplify the numerator.

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

Step 3: Multiply the numerator and denominator by the conjugate of the numerator.

$g'(x) = \lim_{{h \to 0}} \frac{\left( \sqrt{x+h} - \sqrt{x} \right) \cdot \left( \sqrt{x+h} + \sqrt{x} \right)}{h \cdot \left( \sqrt{x+h} + \sqrt{x} \right)}$

Step 4: Apply the difference of squares in the numerator.

$g'(x) = \lim_{{h \to 0}} \frac{(x+h) - x}{h \cdot \left( \sqrt{x+h} + \sqrt{x} \right)}$

$g'(x) = \lim_{{h \to 0}} \frac{h}{h \cdot \left( \sqrt{x+h} + \sqrt{x} \right)}$

Step 5: Cancel out $h$ in the numerator and denominator.

$g'(x) = \lim_{{h \to 0}} \frac{1}{\sqrt{x+h} + \sqrt{x}}$

Step 6: Apply the limit as $h \to 0$.

$g'(x) = \frac{1}{\sqrt{x} + \sqrt{x}} = \frac{1}{2\sqrt{x}}$

Final Answer:

$g'(x) = \frac{1}{2\sqrt{x}}$

Let me know if you need more details or have any questions!

Here are five related questions to explore further:

1. How can we apply the limit process to differentiate other radical functions?
2. What does the derivative of $g(x) = \sqrt{x} - 2$ represent graphically?
3. Can you find the second derivative $g''(x)$ of the given function?
4. How would we differentiate $g(x) = \sqrt{x+1} - 2$ using the limit process?
5. How can we interpret the behavior of $g'(x)$ as $x \to 0$ and $x \to \infty$?

Tip: When dealing with square roots, multiplying by the conjugate is a powerful technique to simplify expressions, especially when using limits.