Answer: [tex]\dfrac{1}{2\sqrt x}[/tex]
Step-by-step explanation:
[tex]\lim_{h \to 0} f(x)=\dfrac{f(x+h)-f(x)}{h}[/tex]
f(x) = [tex]\sqrt x\\[/tex]
f(x+h) = [tex]\sqrt{x+h}[/tex]
[tex]\lim_{h \to 0} f(x)=\dfrac{\sqrt{x+h}-\sqrt x}{h}[/tex]
[tex]=\dfrac{\sqrt{x+h}-\sqrt x}{h}\bigg(\dfrac{\sqrt{x+h}+\sqrt x}{\sqrt{x+h}+\sqrt x}\bigg)[/tex]
[tex]=\dfrac{(x + h)-(x)}{h(\sqrt{x+h}+\sqrt x)}[/tex]
[tex]=\dfrac{h}{h(\sqrt{x+h}+\sqrt x)}[/tex]
[tex]=\dfrac{1}{\sqrt{x+h}+\sqrt x}[/tex]
[tex]=\dfrac{1}{\sqrt{x+0}+\sqrt x}[/tex]
[tex]=\dfrac{1}{2\sqrt x}[/tex]
