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.

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
$$
\begin{align*}
\lim_{x\to 3} \frac{x^{2}+\sqrt{x+6}-12}{x^{2}-9}
&= 1+ \lim_{x\to 3} \frac{\sqrt{x+6}-3}{(x-3)(x+3)}= 1+\frac{1}{6}\lim_{x\to 3} \frac{\sqrt{x+6}-3}{x-3}
\\&= 1+\frac{1}{36} = \boxed{\frac{37}{36}}
\end{align*}$$
