Question

помогите! срочно! даю 50 баллов​

Answer 1

Ответ:

$1)\,0\\\\2)\,-3\\\\3)\,-\frac{1}{16}\\\\4)\,-\frac{145}{42}$

Объяснение:

$1)\,\lim_{x\to\infty}\frac{2x+3}{x^2-9x+8}=\lim_{x\to\infty}\frac{x}{x}\cdot\frac{2+\frac{3}{x}}{x-9+\frac{8}{x}}=\lim_{x\to\infty}\frac{2+0 }{x-9+0}=\lim_{x\to\infty}\frac{2 }{x-9}=0\\-----------------------------------\\\\2)\,\lim_{x \to -2} \frac{x^3+8}{x^2-4}=\lim_{x \to -2}\frac{(x+2)(x^2-2x+4)}{(x+2)(x-2)}=\lim_{x\to-2} \frac{x^2-2x+4}{x-2}=\\\\=\frac{(-2)^2-2(-2)+4}{-2-2}=\frac{4+4+4}{-4}=-3\\-----------------------------------$

$3)\,\lim_{x \to 3} \frac{\sqrt{x+13}-2\sqrt{x+1}}{x^2-9}=\lim_{x \to 3} \frac{(\sqrt{x+13}-2\sqrt{x+1})(\sqrt{x+13}+2\sqrt{x+1})}{(x^2-9)(\sqrt{x+13}+2\sqrt{x+1})}=\\\\=\lim_{x\to3}\frac{x+13-4(x+1)}{(x^2-9)(\sqrt{x+13}+2\sqrt{x+1})}=\lim_{x\to3} \frac{-3x+9}{(x-3)(x+3)(\sqrt{x+13}+2\sqrt{x+1})}=$

$=\lim_{x\to3}\frac{-3(x-3)}{(x-3)(x+3)(\sqrt{x+13}+2\sqrt{x+1})}=\lim_{x\to3}\frac{-3}{((x+3)(\sqrt{x+13}+2\sqrt{x+1})}=\\\\=\frac{-3}{((3+3)(\sqrt{3+13}+2\sqrt{3+1})}=\frac{-3}{(6(4+2\cdot2)}=-\frac{1}{16}\\-----------------------------------\\\\4)\, \lim_{x\to0,4}\frac{5x^3-2x^2+5x-2}{5x^4-2x^3-5x^2+2x}=\lim_{x\to0,4}\frac{5x^3-2x^2+5x-2}{x(5x^3-2x^2-5x+2)}=\\\\=\lim_{x\to0,4}\frac{x^2(5x-2)+(5x-2)}{x(x^2(5x-2)-(5x-2))}=\lim_{x\to0,4}\frac{(x^2+1)(5x-2)}{x(x^2-1)(5x-2)}\, =$

$=\lim_{x\to0,4}\frac{(x^2+1)}{x(x^2-1)}=\frac{((0,4)^2+1)}{0,4((0,4)^2-1)}=-\frac{1,16}{0,4\cdot0,84}= -\frac{1,16}{0,336}=\\\\=-\frac{1160}{336}=-\frac{8\cdot 145}{8\cdot 42}=-\frac{145}{42}$