<s>[INST] Solving this limit $\lim_{x\to 3}\frac{x^2-9-3+\sqrt{x+6}}{x^2-9}$. The question is $\lim_\limits{x\to 3}\frac{x^2-9-3+\sqrt{x+6}}{x^2-9}$.
I hope you guys understand why I have written the numerator like that. So my progress is nothing but $1+\frac{\sqrt{x+6}-3}{x^2-9}$.
Now how do I rationalize the numerator?
It is giving the $\frac{0}{0}$ form after plugging in $3$.
 [/INST] The solution goes as follows:
$$\lim_\limits{x\to 3}\frac{x^2-9-3+\sqrt{x+6}}{x^2-9}$$
$$=\lim_\limits{x\to 3}\left(1+\frac{\sqrt{x+6}-3}{x^2-9}\right)$$
$$=\lim_\limits{x\to 3}\left[1+\frac{(\sqrt{x+6}-3)(\sqrt{x+6}+3)}{(x^2-9)(\sqrt{x+6}+3)}\right]$$
$$=\lim_\limits{x\to 3}\left[1+\frac{x+6-9}{(x^2-9)(\sqrt{x+6}+3)}\right]$$
$$=\lim_\limits{x\to 3}\left[1+\frac{x-3}{(x-3)(x+3)(\sqrt{x+6}+3)}\right]$$
$$=\lim_\limits{x\to 3}\left[1+\frac{1}{(x+3)(\sqrt{x+6}+3)}\right]$$
$$=1+\frac{1}{(3+3)(\sqrt{3+6}+3)}$$
$$=\frac{37}{36}$$
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