<s>[INST] Evaluating $\lim_{x\to-1}\frac{2x^2+5x+3}{2-\sqrt{2+\sqrt{3-x}}}$ 
What is the value of the limit:
$$\lim_{x\to-1}\frac{2x^2+5x+3}{2-\sqrt{2+\sqrt{3-x}}}$$
$1)8\quad\quad\quad\quad\quad2)12\quad\quad\quad\quad\quad3)16\quad\quad\quad\quad\quad4)24$

I evaluated the limit by L'Hopital rule:
$$\lim_{x\to-1}\dfrac{4x+5}{\dfrac1{2\sqrt{3-x}\times2\sqrt{2+\sqrt{3-x}}}}=\dfrac{1}{\dfrac{1}{16}}=16$$
But I wonder is there other approach to evaluate the limit?
 [/INST] HINT
What about multiplying by the conjugate twice?
\begin{align*}
\frac{1}{2 - \sqrt{2+\sqrt{3-x}}} = \frac{2 + \sqrt{2+\sqrt{3-x}}}{2 - \sqrt{3-x}} 
\end{align*}
Similarly, we have
\begin{align*}
\frac{1}{2-\sqrt{3-x}} = \frac{2 + \sqrt{3-x}}{1 + x}
\end{align*}
Finally, notice that
\begin{align*}
2x^{2} + 5x + 3 & = (2x^{2} + 2x) + (3x + 3)\\\\
& = 2x(x+1) + 3(x+1)\\\\
& = (2x+3)(x+1)
\end{align*}
Can you take it from here?
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