#### Answer

The equation of the tangent line $l$ of the curve at point $A$ is $$(l): y=\frac{1}{2}x+\frac{1}{2}$$

#### Work Step by Step

$$y=f(x)=\sqrt x$$ Given point $A (1,1)$
According to definition, the slope of the tangent line $l$ at the given point $A$ is $$m_l=\lim\limits_{x\to1}\frac{f(x)-f(1)}{x-1}$$$$m_l=\lim\limits_{x\to1}\frac{\sqrt x -1}{x-1}$$$$m_l=\lim\limits_{x\to1}\frac{\sqrt x -1}{(\sqrt x -1)(\sqrt x +1)}$$$$m_l=\lim\limits_{x\to1}\frac{1}{\sqrt x +1}$$$$m_l=\frac{1}{\sqrt 1 +1}$$$$m_l=\frac{1}{2}$$
So, the tangent line $l$ of the given function at the given point $A$ would have the form $$(l): y=\frac{1}{2}x+b$$
Since the tangent line $l$ passes through point $A(1,1)$, we have $$\frac{1}{2}\times1+b=1$$$$\frac{1}{2}+b=1$$$$b=\frac{1}{2}$$
Therefore, the equation of the tangent line $l$ of the curve at point $A$ is $$(l): y=\frac{1}{2}x+\frac{1}{2}$$