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

In the math clinic I work at, somebody in a Calculus 1 class asked for help with this limit problem. They have not covered basic differentiation techniques yet, let alone l'Hôpital's rule.

$$\lim_{x\to1}\frac{\sqrt{5-x}-2}{\sqrt{2-x}-1}$$

We have tried various algebraic techniques, such as multiplying the top and bottom of the fraction by the conjugate of the denominator, but haven't had any success at getting rid of the indeterminate form.

How can this limit be solved, using only techniques that would be available to a beginning Calculus 1 student?

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