Question

помогите пожалуйста
Найти пределы функций:
$1) \lim_{x \to \infty} \frac{3x^2-4x-2}{9x^2+x-2 } 2) \lim_{x \to \+1} \frac{x^2-1}{X^2+6-7} 3) \lim_{x \to \infty} \frac{3x^2+4x-2}{6x^2+x-5} 4) \lim_{x \to \+1} \frac{x^2-1}{x^2+2x-3}$

Answer 1

$1)\; \; \lim\limits _{x \to \infty} \frac{3x^2-4x-2}{9x^2+x-2}= \lim\limits _{x \to \infty}\, \frac{3-\frac{4}{x}-\gfrac{2}{x^2}}{9+\frac{1}{x^2}-\frac{2}{x^2}} = \frac{3}{9}=\frac{1}{3}\\\\2)\; \; \lim\limits _{x \to 1}\frac{x^2-1}{x^2+6x-7} = \lim\limits _{x \to 1} \frac{(x-1)(x+1)}{(x-1)(x+7)} = \lim\limits _{x \to 1} \frac{x+1}{x+7} = \frac{1+1}{1+7} =\frac{2}{8}= \frac{1}{4}$

$3)\; \; \lim\limits _{x \to \infty} \frac{3x^2+4x-2}{6x^2+x-5} = \lim\limits _{x \to \infty} \frac{3+\frac{4}{x}-\frac{2}{x^2}}{6+\frac{1}{x}-\frac{5}{x^2}}= \frac{3}{6}= \frac{1}{2} \\\\4)\; \; \lim\limits _{x \to 1} \frac{x^2-1}{x^2+2x-3} = \lim\limits _{x \to 1} \frac{(x-1)(x+1)}{(x-1)(x+3)} = \lim\limits _{x \to 1} \frac{x+1}{x+3} = \frac{2}{4}= \frac{1}{2}$