How can I evaluate $\lim_{x\to1}\frac{\sqrt{5-x}-2}{\sqrt{2-x}-1}$ without invoking l'Hôpital's rule?

$$\begin{align} & \hphantom{=} \lim_{x\to1}\frac{\sqrt{5-x}-2}{\sqrt{2-x}-1} \\ & = \lim_{x\to 1} \frac{\sqrt{5-x}-2}{\sqrt{2-x}-1}\frac{\sqrt{2-x}+1}{\sqrt{2-x}+1} \\ & = \lim_{x\to 1} \frac{(\sqrt{5-x}-2)(\sqrt{2-x}+1)}{1-x} \\ & = \lim_{x\to 1} \frac{(\sqrt{5-x}-2)(\sqrt{2-x}+1)}{1-x}\frac{\sqrt{5-x}+2}{\sqrt{5-x}+2} \\ & = \lim_{x\to 1} \frac{(1-x)(\sqrt{2-x}+1)}{1-x}\frac{1}{\sqrt{5-x}+2} \\ & = \lim_{x\to 1} \frac{\sqrt{2-x}+1}{\sqrt{5-x}+2} \\ & = \frac{1}{2} \end{align}$$

When in doubt, multiply by the "conjugate".

$$\lim_{x\to1}\frac{\sqrt{5-x}-2}{\sqrt{2-x}-1} = \lim_{x\to1}\frac{\frac{\sqrt{5-x}-2}{1-x}}{\frac{\sqrt{2-x}-1}{1-x}}$$ $$= \lim_{x\to1}\frac{\frac{\sqrt{5-x}-2}{5-x-4}}{\frac{\sqrt{2-x}-1}{2-x-1}}= \lim_{x\to1}\frac{\frac{\sqrt{5-x}-2}{(\sqrt{5-x}-{2})*(\sqrt{5-x}+{2})}}{\frac{\sqrt{2-x}-1}{{(\sqrt{1-x}-{1})*(\sqrt{2-x}+{1})}}}=\lim_{x\to1}\frac{\sqrt{2-x}+1}{\sqrt{5-x}+2}=\frac{1}{2} $$