Limit (trig) without L'Hospitals's Hello while I was studying, I found some difficulties solving this limit,specifically how to get rid of the $\sin$
$$\lim_{x\to3}\biggl(\biggl(\frac{1}{\sqrt{x+3}}-\frac{1}{\sqrt{6}}\biggr)\biggl(\frac{x+3}{\sin({2x-6})}\biggr)\biggr)$$
I am supposed to solve this without the use of L'Hospital's, and so far I tried,trying to get the $\sin$ to a form $\frac{\sin(x)}{x} $ where $x=2x+6$ but I still get the indeterminate form $0\times\infty$
Also tried turning it to $1-\cos^2(2x-6)$ yet i realized it didn't change anything
And lastly I wanted to make the silly attempt to make $\sin(x)=x$ but I am almost certain i am not able to do that
Any help or advice on how to deal with this limit or any other similar is much appreciated,I'm not interested on the result itself but more on how to proceed should I encounter similar ones.
I hope i made myself clear and thanks in advance!

It might be useful to set $t=x-3$ and consider $t\to 0$.
\begin{eqnarray*} \left(\frac{1}{\sqrt{x+3}}-\frac{1}{\sqrt{6}}\right)\frac{x+3}{\sin({2x-6})}
& \stackrel{t=x-3}{=} & \left(\frac{1}{\sqrt{t+6}}- \frac 1{\sqrt 6}\right)\frac{t+6}{\sin(2t)} \\
& = & \left(\frac{\sqrt 6 - \sqrt{t+6}}{\sqrt 6}\right)\frac{\sqrt{t+6}}{\sin(2t)} \\
& = & -\frac{1}{2\sqrt 6}\frac{\sqrt{t+6}}{\sqrt 6 + \sqrt{t+6}}\frac{2t}{\sin (2t)} \\
&\stackrel{t\to 0}{\longrightarrow} & -\frac 1{4\sqrt 6}
\end{eqnarray*}
