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 $3$.
 A: Hint. You may write, as $x \to 3$,
$$
\frac{\sqrt{x+6}-3}{x^2-9}=\frac{(x+6)-9}{(x-3)(x+3)(\sqrt{x+6}+3)}=\frac1{(x+3)(\sqrt{x+6}+3)}
$$
A: The solution goes as follows:
$$\lim_\limits{x\to 3}\frac{x^2-9-3+\sqrt{x+6}}{x^2-9}$$
$$=\lim_\limits{x\to 3}\left(1+\frac{\sqrt{x+6}-3}{x^2-9}\right)$$
$$=\lim_\limits{x\to 3}\left[1+\frac{(\sqrt{x+6}-3)(\sqrt{x+6}+3)}{(x^2-9)(\sqrt{x+6}+3)}\right]$$
$$=\lim_\limits{x\to 3}\left[1+\frac{x+6-9}{(x^2-9)(\sqrt{x+6}+3)}\right]$$
$$=\lim_\limits{x\to 3}\left[1+\frac{x-3}{(x-3)(x+3)(\sqrt{x+6}+3)}\right]$$
$$=\lim_\limits{x\to 3}\left[1+\frac{1}{(x+3)(\sqrt{x+6}+3)}\right]$$
$$=1+\frac{1}{(3+3)(\sqrt{3+6}+3)}$$
$$=\frac{37}{36}$$
A: 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
$$\lim_{y\to0}\dfrac1{y(y+6)^2+6(y+6)}=?$$
A: Your first step is good. Now you want to compute
$$
\lim_{x\to 3}\frac{\sqrt{x+6}-3}{x-3}
$$
because the factor $x+3$ at the denominator poses no problem. This should remind you of the definition of derivative and indeed it's the derivative of $f(x)=\sqrt{x+6}$ at $3$. Since
$$
f'(x)=\frac{1}{2\sqrt{x+6}}
$$
for $x>-6$, you have $f'(3)=1/6$ and so
$$
\lim_{x\to 3}\frac{x^2-9-3+\sqrt{x+6}}{x^2-9}=
\lim_{x\to 3}\left(1+\frac{\sqrt{x+6}-3}{x-3}\frac{1}{x+3}\right)=
1+\frac{1}{6}\frac{1}{6}
$$
If you don't know about derivatives, just do
$$
\lim_{x\to 3}\frac{\sqrt{x+6}-3}{x-3}=
\lim_{x\to3}\frac{(\sqrt{x+6}-3)(\sqrt{x+6}+3)}{(x-3)(\sqrt{x+6}+3)}
=\lim_{x\to3}\frac{x-3}{(x-3)(\sqrt{x+6}+3)}=\dots
$$
