Answer
Yes, $y=\frac{1}{2}x+\frac{1}{2}$ at $(1, 1)$.
Work Step by Step
Step 1. Given $y=f(x)=\sqrt x$, we have
$y'=f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}\frac{\sqrt {x+h}-\sqrt x}{h}
=\lim_{h\to0}\frac{ (\sqrt {x+h}-\sqrt x) (\sqrt {x+h}+\sqrt x) }{h (\sqrt {x+h}+\sqrt x)}=
\lim_{h\to0}\frac{ x+h-x }{h (\sqrt {x+h}+\sqrt x)}=\frac{ 1 }{ (\sqrt {x+0}+\sqrt x)}=\frac{1}{2\sqrt x}$
Step 2. Assume a tangent line to the curve crosses the x-axis at $x=-1$ which means that point $(-1,0)$ is on the line; we can get the line equation as $y=m(x+1)$, where $m$ is the slope given by $f'(x)$.
Step 3. The intersection of the line and the curve can be found as $m(x+1)=\sqrt x$ or $\frac{1}{2\sqrt x}(x+1)=\sqrt x$, which gives $x+1=2x$ or $x=1$ and the point of intersection is then $(1, 1)$
Step 4. With the slope $m=\frac{1}{2\sqrt x}=\frac{1}{2}$, the equation of the tangent line can be written as $y-1=\frac{1}{2}(x-1)$, which gives $y=\frac{1}{2}x+\frac{1}{2}$ at the point of tangency of $(1, 1)$.
