Answer:
[tex]\dfrac{1}{2\sqrt{x}}[/tex]
Step-by-step explanation:
[tex]f(x) = \sqrt{x} = x^{\frac{1}{2}}[/tex]
[tex]f(x+h) = \sqrt{x+h} = (x+h)^{\frac{1}{2}}[/tex]
We use binomial expansion for [tex] (x+h)^{\frac{1}{2}}[/tex]
This can be rewritten as
[tex][x(1+\dfrac{h}{x})]^{\frac{1}{2}}[/tex]
[tex]x^{\frac{1}{2}}(1+\dfrac{h}{x})^{\frac{1}{2}}[/tex]
From the expansion
[tex](1+x)^n=1+nx+\dfrac{n(n-1)}{2!}+\ldots[/tex]
Setting [tex]x=\dfrac{h}{x}[/tex] and [tex]n=\frac{1}{2}[/tex],
[tex](1+\dfrac{h}{x})^{\frac{1}{2}}=1+(\dfrac{h}{x})(\dfrac{1}{2})+\dfrac{\frac{1}{2}(1-\frac{1}{2})}{2!}(\dfrac{h}{x})^2+\tldots[/tex]
[tex]=1+\dfrac{h}{2x}-\dfrac{h^2}{8x^2}+\ldots[/tex]
Multiplying by [tex]x^{\frac{1}{2}}[/tex],
[tex]x^{\frac{1}{2}}(1+\dfrac{h}{x})^{\frac{1}{2}}=x^{\frac{1}{2}}+\dfrac{h}{2x^{\frac{1}{2}}}-\dfrac{h^2}{8x^{\frac{3}{2}}}+\ldots[/tex]
[tex]x^{\frac{1}{2}}(1+\dfrac{h}{x})^{\frac{1}{2}}-x^{\frac{1}{2}}=\dfrac{h}{2x^{\frac{1}{2}}}-\dfrac{h^2}{8x^{\frac{3}{2}}}+\ldots[/tex]
[tex]\dfrac{x^{\frac{1}{2}}(1+\dfrac{h}{x})^{\frac{1}{2}}-x^{\frac{1}{2}}}{h}=\dfrac{1}{2x^{\frac{1}{2}}}-\dfrac{h}{8x^{\frac{3}{2}}}+\ldots[/tex]
The limit of this as [tex]h\to 0[/tex] is
[tex]\lim_{h\to0} \dfrac{f(x+h)-f(x)}{h}=\dfrac{1}{2x^{\frac{1}{2}}}=\dfrac{1}{2\sqrt{x}}[/tex] (since all the other terms involve [tex]h[/tex] and vanish to 0.)
