<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] Hint. You may write, as $x \to 3$,
$$
\frac{\sqrt{x+6}-3}{x^2-9}=\frac{(x+6)-9}{(x-3)(x+3)(\sqrt{x+6}+3)}=\frac1{(x+3)(\sqrt{x+6}+3)}
$$
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