[tex]\displaystyle\lim_{x\to9}\frac{\sqrt x-3}{x-9}=\lim_{x\to9}\frac{\sqrt x-3}{(\sqrt x-3)(\sqrt x+3)}=\lim_{x\to9}\frac1{\sqrt x+3}=\frac1{\sqrt9+3}=\frac16[/tex]
Alternatively, you can observe that if [tex]f(x)=\sqrt x[/tex], then by the definition of the derivative, you have
[tex]f'(9)=\displaystyle\lim_{x\to9}\frac{f(x)-f(9)}{x-9}[/tex]
You have
[tex]f(x)=\sqrt x\implies f'(x)=\dfrac1{2\sqrtx}\implies f'(9)=\dfrac1{2\sqrt9}=\dfrac16[/tex]
