Ответ:

[tex]\displaystyle \frac{1}{2}[/tex]

Объяснение:

[tex]\displaystyle \lim_{x\to 1 } \frac{\sqrt{5-x}-2}{\sqrt{2-x}-1}= \left[\frac{0}{0} \right] =\lim_{x\to 1 } \frac{(\sqrt{5-x}-2)'}{(\sqrt{2-x}-1)'}=\\\\\lim_{x\to 1 } \frac{(\sqrt{5-x})'-2'}{(\sqrt{2-x)})'-1'}=\lim_{x\to 1 } \frac{\frac{1}{2\sqrt{5-x}}\cdot(5-x)'}{\frac{1}{2\sqrt{2-x}}\cdot(2-x)'}=[/tex]

[tex]\displaystyle\lim_{x\to 1 } \frac{-\frac{1}{2\sqrt{5-x}}}{-\frac{1}{2\sqrt{2-x}}}=\lim_{x\to 1 } \frac{2\sqrt{2-x}}{2\sqrt{5-x}}=\\\\ \lim_{x\to 1 } \frac{\sqrt{2-x}}{\sqrt{5-x}}=\frac{\sqrt{2-1}}{\sqrt{5-1}}=\frac{\sqrt 1}{\sqrt 4}=\frac{1}{2}[/tex]