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.
 A: An elementary approach: "recognize the derivative."
You can rewrite, for $x\notin\{-3,3\}$,
$$
\frac{x^{2}+\sqrt{x+6}-12}{x^{2}-9}
 = 1+  \frac{\sqrt{x+6}-3}{x^{2}-9}
 = 1+  \frac{\sqrt{x+6}-3}{(x-3)(x+3)}
 = 1+  \frac{\sqrt{x+6}-3}{(x-3)(x+3)} \tag{1}
$$
so the question boils down to computing $\lim_{x\to 3} \frac{\sqrt{x+6}-3}{x-3}$  (the rest is "under control").
Now, we can recognize a derivative here: let $f\colon[0,\infty) \to [0,\infty)$ be defined by 
$$
f(x) = \sqrt{x+6}, \qquad x\geq 0\,. \tag{2}
$$
Note that $f$ is differentiable, with $f'(x) = \frac{1}{2\sqrt{x+6}}$. Therefore,
$$
\frac{1}{6} = \frac{1}{2\sqrt{9}} = f'(3) = \lim_{x\to 3} \frac{f(x)-f(3)}{x-3} = \lim_{x\to 3} \frac{\sqrt{x+6}-3}{x-3} \tag{3}
$$
giving the desired limit.
Putting it together, from (1) and (3) we get
$$
\begin{align*}
\lim_{x\to 3} \frac{x^{2}+\sqrt{x+6}-12}{x^{2}-9}
&= 1+ \lim_{x\to 3} \frac{\sqrt{x+6}-3}{(x-3)(x+3)}= 1+\frac{1}{6}\lim_{x\to 3} \frac{\sqrt{x+6}-3}{x-3}
\\&= 1+\frac{1}{36} = \boxed{\frac{37}{36}}
\end{align*}$$
A: $$\frac{x^{2}+\sqrt{x+6}-12}{x^{2}-9}
 = 1+  \frac{\sqrt{x+6}-3}{(x-3)(x+3)}=1+\frac{1}{(x+3)(\sqrt{x+6}+3)}$$
A: $$\lim_{x\to 3} \frac{x^{2}+\sqrt{x+6}-12}{x^{2}-9} $$
$$ = \lim_{x\to 3} \frac{x^{2}+\sqrt{x+6}-9-3}{x^{2}-9}$$
$$ = \lim_{x\to 3} \frac{x^{2}-9+\sqrt{x+6}-3}{x^{2}-9}$$
$$ = \lim_{x\to 3} \Big(\frac{x^{2}-9}{x^{2}-9}+\frac{\sqrt{x+6}-3}{x^{2}-9}\Big) $$
$$ = \lim_{x\to 3} \frac{x^{2}-9}{x^{2}-9} + \lim_{x\to 3} \frac{\sqrt{x+6}-3}{x^{2}-9}$$
$$ = \lim_{x\to 3} 1 + \lim_{x\to 3} \frac{\sqrt{x+6}-3}{x^{2}-9}$$
$$ = 1 + \lim_{x\to 3} \frac{\sqrt{x+6}-3}{x^{2}-9}$$
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 $$\lim_{x\to 3} \frac{x^{2}+\sqrt{x+6}-12}{x^{2}-9} = 1 + \lim_{x\to 3} \frac{\sqrt{x+6}-3}{x^{2}-9} = 1 + \frac{1}{36} = \frac{36}{36} + \frac{1}{36} $$
$$= \frac{37}{36}$$
