Question

ДАЮ 20 БАЛЛОВ, РЕШИТЕ ПОЖАЛУЙСТА, ЛЮБЫЕ

Answer 1

1)
$\lim_{x \to \infty} \frac{5x^2+11x-3}{-x^2-2x+3} = \lim_{x \to \infty} \frac{ \frac{5x^2}{x^2} + \frac{11x}{x^2} - \frac{3}{x^2} }{- \frac{x^2}{x^2} - \frac{2x}{x^2} + \frac{3}{x^2} } = \frac{5}{-1} =-5$

2)
$\lim_{x \to \infty} \frac{2x^3-2x^2}{x^4+x-3} = \lim_{x \to \infty} \frac{ \frac{2x^3}{x^3} - \frac{2x^2}{x^3} }{ \frac{x^4}{x^3} + \frac{x}{x^3} - \frac{3}{x^3} } = \lim_{x \to \infty} \frac{2}{x} = \frac{2}{\infty}=0$

3)
$\lim_{x \to -2} \frac{x+2}{x^2-x-6} = \lim_{x \to -2} \frac{x+2}{(x+2)(x-3)} =\lim_{x \to -2} \frac{1}{x-3} = \frac{1}{-5} =-0.2$

4)
$\lim_{x \to 3} \frac{9-x^2}{27-x^3} = \lim_{x \to 3} \frac{(3-x)(3+x)}{(3-x)(x^2+3x+9)} = \lim_{x \to 3} \frac{3+x}{x^2+3x+9} = \frac{6}{27}= \frac{2}{9}$

5)

6)
$\lim_{x \to 4} \frac{x-4}{ \sqrt{x+5} -3} = \lim_{x \to 4} \frac{(x-4)(\sqrt{x+5} +3)}{ (\sqrt{x+5} -3)(\sqrt{x+5} +3)} =\\\\ \lim_{x \to 4} \frac{(x-4)(\sqrt{x+5} +3)}{ x-4} = lim_{x \to 4} (\sqrt{x+5} +3)= 6$

7)
$\lim_{x \to 0} \frac{ \sqrt{1+x}-1 }{5x} = \lim_{x \to 0} \frac{ (\sqrt{1+x}-1)(\sqrt{1+x}+1) }{5x(\sqrt{1+x}+1)} =\\\\\lim_{x \to 0} \frac{ x}{5x(\sqrt{1+x}+1)} =\lim_{x \to 0} \frac{ 1}{5(\sqrt{1+x}+1)} = \frac{1}{5*2} =0.1$