Question

1. (10 points) if $g(x) = \sqrt{x}$, find $g(x)$ (using the limit definition of derivative).

Explanation:

Step1: Recall limit definition of derivative

The limit definition is $g'(x) = \lim_{h \to 0} \frac{g(x+h) - g(x)}{h}$. For $g(x)=\sqrt{x}$, substitute:
$\lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h}$

Step2: Rationalize the numerator

Multiply numerator and denominator by $\sqrt{x+h} + \sqrt{x}$:
$\lim_{h \to 0} \frac{(\sqrt{x+h} - \sqrt{x})(\sqrt{x+h} + \sqrt{x})}{h(\sqrt{x+h} + \sqrt{x})} = \lim_{h \to 0} \frac{(x+h) - x}{h(\sqrt{x+h} + \sqrt{x})}$

Step3: Simplify the expression

Cancel $x$ in numerator: $\lim_{h \to 0} \frac{h}{h(\sqrt{x+h} + \sqrt{x})} = \lim_{h \to 0} \frac{1}{\sqrt{x+h} + \sqrt{x}}$

Step4: Evaluate the limit

Substitute $h=0$: $\frac{1}{\sqrt{x} + \sqrt{x}} = \frac{1}{2\sqrt{x}}$

Answer:

$\frac{1}{2\sqrt{x}}$