НАЙДИТЕ ПРЕДЕЛЫ.ДАЮ 40 БАЛЛОВ

Решите задачу:

$1)\; \; lim_{x\to 1} \frac{\sqrt{2-x}-1}{\sqrt{5-x}-2} = \lim_{x \to 1} \frac{(\sqrt{2-x}-1)(\sqrt{2-x}+1)(\sqrt{5-x}+2)}{(\sqrt{5-x}-2)(\sqrt{5-x}+2)(\sqrt{2-x}+1)} =\\\\= \lim_{x \to 1} \frac{(2-x-1)(\sqrt{5-x}+2)}{(5-x-4)(\sqrt{2-x}+1)} =lim_{x\to 1} \frac{(1-x)(\sqrt{5-x}+2)}{(1-x)(\sqrt{2-x}+1)} =\frac{\sqrt{5-1}+2}{\sqrt{2-1}+1}=\frac{4}{2}=2\\\\2)\; \; \lim_{x \to 0} \frac{tg2x}{sin5x}=lim_{x\to 0} (\frac{tg2x}{2x}\cdot 2x\cdot \frac{5x}{sin5x}\cdot \frac{1}{5x})=$

$=lim_{x\to 0}(1\cdot 2x\cdot 1\cdot \frac{1}{5x})=\frac{2}{5}\\\\3)\; \; \lim_{x \to 1} \frac{x^3-x^2+3x-3}{2x^3-2x^2+x-1} =lim_{x\to 1} \frac{x^2(x-1)+3(x-1)}{2x^2(x-1)+(x-1)} =\\\\=lim_{x\to 1} \frac{(x-1)(x^2+3)}{(x-1)(2x^2+1)} =lim_{x\to 1} \frac{x^2+3}{2x^2+1} =\frac{1+3}{2+1}=\frac{4}{3}$