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.

$$\lim_{x\to 3} \frac{x^{2}+\sqrt{x+6}-12}{x^{2}-9} $$
$$ = \lim_{x\to 3} \frac{x^{2}+\sqrt{x+6}-9-3}{x^{2}-9}$$
$$ = \lim_{x\to 3} \frac{x^{2}-9+\sqrt{x+6}-3}{x^{2}-9}$$
$$ = \lim_{x\to 3} \Big(\frac{x^{2}-9}{x^{2}-9}+\frac{\sqrt{x+6}-3}{x^{2}-9}\Big) $$
$$ = \lim_{x\to 3} \frac{x^{2}-9}{x^{2}-9} + \lim_{x\to 3} \frac{\sqrt{x+6}-3}{x^{2}-9}$$
$$ = \lim_{x\to 3} 1 + \lim_{x\to 3} \frac{\sqrt{x+6}-3}{x^{2}-9}$$
$$ = 1 + \lim_{x\to 3} \frac{\sqrt{x+6}-3}{x^{2}-9}$$
After rationalizing the numerator to remove the radical:
$$\lim_{x\to 3}\frac{\sqrt{x+6}-3}{x^{2}-9} \cdot \frac{\sqrt{x+6}+3}{\sqrt{x+6}+3} = \lim_{x\to 3}\frac{x+6-9}{(x^{2}-9)(\sqrt{x+6}+3)} = \lim_{x\to 3}\frac{x-3}{(x^{2}-9)(\sqrt{x+6}+3)} $$
After factoring $ x^2 - 9 $, simplifying, and direct substitution:
$$ \lim_{x\to 3}\frac{x-3}{(x^{2}-9)(\sqrt{x+6}+3)} = \lim_{x\to 3}\frac{x-3}{(x-3)(x+3)(\sqrt{x+6}+3)} = \lim_{x\to 3}\frac{1}{(x+3)(\sqrt{x+6}+3)} = \frac{1}{6\cdot 6} = \frac{1}{36}$$
Thus $$\lim_{x\to 3} \frac{x^{2}+\sqrt{x+6}-12}{x^{2}-9} = 1 + \lim_{x\to 3} \frac{\sqrt{x+6}-3}{x^{2}-9} = 1 + \frac{1}{36} = \frac{36}{36} + \frac{1}{36} $$
$$= \frac{37}{36}$$
