Can you help me with it and explain the steps
$$\lim_{x \to 1} \frac{\sqrt{5-x}-2}{\sqrt{2-x}-1}$$
I tried to multiply at conjugate expression but I failed.
Hint: $$\frac{\sqrt{5-x}-2}{\sqrt{2-x}-1}=\frac{(\sqrt{5-x}-2)(\sqrt{5-x}+2)(\sqrt{2-x}+1)}{(\sqrt{2-x}-1)(\sqrt{2-x}+1)(\sqrt{5-x}+2)}.$$
You could multiply by $(\sqrt{5-x}+2)(\sqrt{2-x}+1)$ both the numerator and denominator of this fraction.
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Hint: $$\frac{\sqrt{5-x}-2}{\sqrt{2-x}-1}=\frac{(\sqrt{5-x}-2)(\sqrt{5-x}+2)(\sqrt{2-x}+1)}{(\sqrt{2-x}-1)(\sqrt{2-x}+1)(\sqrt{5-x}+2)}.$$