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 $3$.

Set $\sqrt{x+6}-3=y\implies x=(y+3)^2-6=y^2+6y+3$
$$\implies\lim_{x\to3}\dfrac{\sqrt{x+6}-3}{x^2-9}$$
$$=\lim_{y\to0}\dfrac y{(y^2+6y+3)^2-9}$$
$$=\lim_{y\to0}\dfrac y{(y^2+6y)^2+6(y^2+6y)}$$
As $y\to0,y\ne0$
So, cancel out $y$ from N & D to get
$$\lim_{y\to0}\dfrac1{y(y+6)^2+6(y+6)}=?$$
