Evaluate the limit: $\lim_{x\to 3} \frac{x^{2}+\sqrt{x+6}-12}{x^{2}-9}$ $$\lim_{x\to 3} \frac{x^{2}+\sqrt{x+6}-12}{x^{2}-9} $$ I want to know how to evaluate without using L'Hopital Rule. I'm unable to factorise or simplify it suitably.

$$\frac{x^{2}+\sqrt{x+6}-12}{x^{2}-9} = 1+ \frac{\sqrt{x+6}-3}{(x-3)(x+3)}=1+\frac{1}{(x+3)(\sqrt{x+6}+3)}$$