Question: Find the limit of $f(x)=\frac{\sqrt{5-x}-1}{2-\sqrt{x}}$ when x approaches 4.

I have no idea where to start with this problem.

Answer: 
**Without using L'Hospital's rule**

Notice, 

Rationalizing the numerator & denominator as follows  

$$\lim\_{x\to 4}\frac{\sqrt{5-x}-1}{2-\sqrt{x}}$$ 
$$=\lim\_{x\to 4}\frac{(\sqrt{5-x}-1)(\sqrt{5-x}+1)(2+\sqrt{x})}{(2-\sqrt{x})(2+\sqrt{x})(\sqrt{5-x}+1)}$$ 
$$=\lim\_{x\to 4}\frac{(5-x-1)(2+\sqrt{x})}{(4-x)(\sqrt{5-x}+1)}$$ 
$$=\lim\_{x\to 4}\frac{(4-x)(2+\sqrt{x})}{(4-x)(\sqrt{5-x}+1)}$$ 
$$=\lim\_{x\to 4}\frac{2+\sqrt{x}}{\sqrt{5-x}+1}$$ 
$$=\frac{2+\sqrt{4}}{\sqrt{5-4}+1}=\frac{4}{2}=\color{blue}{2}$$
Notice, 

Using l'Hopital's rule for $\frac{0}{0}$ form  

$$\lim\_{x\to 4}\frac{\sqrt{5-x}-1}{2-\sqrt{x}}$$ 
$$=\lim\_{x\to 4}\frac{\frac{d}{dx}(\sqrt{5-x}-1)}{\frac{d}{dx}(2-\sqrt{x})}$$ 

$$=\lim\_{x\to 4}\frac{\frac{-1}{2\sqrt{5-x}}}{\frac{-1}{2\sqrt{x}}}$$ $$=\lim\_{x\to 4}\frac{\sqrt x}{\sqrt{5-x}}$$ $$=\frac{\sqrt{4}}{\sqrt{5-4}}=\frac{2}{1}=\color{red}{2}$$