$$(1)\quad f'(3)=\lim_{x\to 3}\frac{f(x)-f(3)}{x-3}=\lim_{x\to 3}\frac{\sqrt x-\sqrt 3}{x-3}
=\lim_{x\to 3}\frac{\sqrt x-\sqrt 3}{x-3}\frac{\sqrt x+\sqrt 3}{\sqrt x+\sqrt 3}
=\lim_{x\to 3}\frac{x-3}{(x-3)(\sqrt x+\sqrt 3)}\\
=\lim_{x\to 3}\frac{1}{\sqrt x+\sqrt 3}=\frac{1}{\sqrt 3+\sqrt 3}=\frac{1}{2\sqrt 3}
$$
$$(2)\quad f'(3)=\lim_{h\to 0}\frac{f(3+h)-f(3)}{h}=\lim_{h\to 0}\frac{\sqrt{3+h}-\sqrt 3}{h}
=\lim_{h\to 0}\frac{\sqrt{3+h}-\sqrt 3}{h}\frac{\sqrt{3+h}+\sqrt 3}{\sqrt{3+h}+\sqrt 3}\\
=\lim_{h\to 0}\frac{(3+h)-3}{h(\sqrt{3+h}+\sqrt 3)}=\lim_{h\to 0}\frac{h}{h(\sqrt{3+h}+\sqrt 3)}
=\lim_{h\to 0}\frac{1}{\sqrt{3+h}+\sqrt 3}=\lim_{h\to 0}\frac{1}{\sqrt 3+\sqrt 3}=\frac{1}{2\sqrt 3}
$$

We see that the derivative of $f$ at $x=3$ is $\frac{1}{2\sqrt 3}$, i.e. $f'(3)=\frac{1}{2\sqrt 3}$.

Just as in the prior slide computing instantaneous velocity, notice that in both computations (until the denominator was cancelled) if we plugged $x=3$ in, we would get the indeterminate form $\frac{0}{0}$ (try it) so we had to do more work.

Also, again notice that the work required in either computation is similar.  Be sure you practice method (2)!