Answer
$$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\frac{\sqrt7}{14}$$
Work Step by Step
$$f(x)=\sqrt x\hspace{1cm} x=7$$
So $f(x+h)=\sqrt{x+h}$
Therefore, $$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt x}{h}$$
-Multiply both numerator and denominator with $\sqrt{x+h}+\sqrt x$, so that
- Numerator: $(\sqrt{x+h}-\sqrt x)(\sqrt{x+h}+\sqrt x)=(x+h)-x=h$
- Denominator: $h(\sqrt{x+h}+\sqrt x)$
That means $$\lim_{h\to0}\frac{f(x+h)f(x)}{h}=\lim_{h\to0}\frac{h}{h(\sqrt{x+h}+\sqrt x)}$$ $$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}\frac{1}{(\sqrt{x+h}+\sqrt x)}$$
- Substitute $x=7$ here: $$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}\frac{1}{\sqrt{7+h}+\sqrt7}$$ $$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\frac{1}{\sqrt{7+0}+\sqrt7}=\frac{1}{\sqrt7+\sqrt7}$$ $$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\frac{1}{2\sqrt7}$$
Multiply both numerator and denominatory with $\sqrt7$: $$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\frac{\sqrt7}{2\times7}=\frac{\sqrt7}{14}$$
