Solve $\lim_{x \rightarrow 1}\frac{\sqrt x -1}{2-\sqrt{x+3}}$ without using L'Hôpital's rule. $\lim_{x \rightarrow 1}\frac{\sqrt x -1}{2-\sqrt{x+3}}$ 
(I have nothing against using L'Hôpital's rule, but my book expects me to solve this without using it which is something I don't quite dominate yet, and so I will try for practice)
I tried the following:
$$\lim_{x \rightarrow 1}\frac{\sqrt x -1}{2-\sqrt{x+3}} \cdot \frac{2+\sqrt{x+3}}{2+\sqrt{x+3}} = \lim_{x \rightarrow 1}\frac{(\sqrt x -1)(2+\sqrt{x+3})}{4-(x+3)} = \lim_{x \rightarrow 1}\frac{2\sqrt x +\sqrt x \sqrt{x+3}-2-\sqrt{x+3}}{1-x} = ???$$
What do I do next?

In your last step, rather than expand out the numerator you should leave it alone, and instead note that
$$ \frac{(\sqrt{x}-1)(2+\sqrt{x+3})}{1-x}=\frac{(\sqrt{x}-1)(2+\sqrt{x+3})}{(1-\sqrt{x})(1+\sqrt{x})}=-\frac{(2+\sqrt{x+3})}{1+\sqrt{x}} $$
