Answer
$a.$ The slope is $m_{tan}=\frac{1}{2}.$
$b.$ The equation is $y=\frac{1}{2}x.$
Work Step by Step
$a.$ Using the formula from definition (2) with $a=2$ and $f(a)=1$ (coordinates of the point $P(2,1)$) we have
$$m_{tan}=\lim_{h\to0}\frac{f(2+h)-f(2)}{h}=\lim_{h\to0}\frac{\sqrt{(2+h)-1}-1}{h}=\lim_{h\to0}\frac{\sqrt{h+1}-1}{h}\cdot\frac{\sqrt{h+1}+1}{\sqrt{h+1}+1}=\lim_{h\to0}\frac{\sqrt{h+1}^2-1}{h(\sqrt{h+1}+1)}=\lim_{h\to0}\frac{h+1-1}{h(\sqrt{h+1}+1)}=\lim_{h\to0}\frac{h}{h(\sqrt{h+1}+1)}=\lim_{h\to0}\frac{1}{\sqrt{h+1}+1}=\frac{1}{\sqrt{0+1}+1} = \frac{1}{2}.$$
$b.$ Using the formula $y-f(a)=m_{tan}(x-a)$ with the same values for $a$ and $f(a)$ as in part $a$ and the calculated value $m_{tan} = 1/2$ we get
$$y-1=\frac{1}{2}(x-2)\Rightarrow y-1=\frac{1}{2}x-1$$ which gives
$$y=\frac{1}{2}x.$$
