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Решите задачу:

$\lim_{x \to \infty} \frac{1-2x}{3x-2} = \lim_{x \to \infty} \frac{x( \frac{1}{x}-2) }{x(3- \frac{2}{x})} = \lim_{x \to \infty} \frac{ \frac{1}{x}-2}{3- \frac{2}{x} }=- \frac{2}{3}$

$\lim_{x \to 0} \frac{ \sqrt{1+x} - \sqrt{1-x} }{3x} = \lim_{x \to 0} \frac{ 1+x - 1+x }{3x( \sqrt{1+x} +\sqrt{1-x})} = \lim_{x \to 0} \frac{2}{ \sqrt{1+x} + \sqrt{1-x}} =1$

$\lim_{x \to \infty} \frac{3x+1}{2x^3+1}= \lim_{x \to \infty} \frac{x(3+ \frac{1}{x} )}{x(2x^2+ \frac{1}{x} )}= \lim_{x \to \infty} \frac{3}{2x^2} =0$

$\lim_{x \to 7} \frac{ \sqrt{2+x} -3}{x-7} = \lim_{x \to 7} \frac{ 2+x -9}{(x-7)(\sqrt{2+x} +3)} = \lim_{x \to 7} \frac{1}{\sqrt{2+x} +3} = \frac{1}{6}$

$\lim_{x \to \infty} \frac{x^3(2+ \frac{1}{x}- \frac{5}{x^3}) }{x^3(1+ \frac{1}{x^2}- \frac{2}{x^3}) } = \lim_{x \to \infty} \frac{2+ \frac{1}{x}- \frac{5}{x^3}}{1+ \frac{1}{x^2}- \frac{2}{x^3}} =2$