Question

Решите пожалуйста)) lim х²+3х-12/-6х²+х-5 x→∞ lim x²-x-6/x²-6x+9 x→3 lim 4x²+11x-3/3x²+10x+3 x→-3 lim x²+2x+1/2x²+x+3 x→∞

Answer 1

$1)\; \; \lim\limits _{x \to \infty}\frac{x^2+3x-12}{-6x^2+x-5}=\lim\limits _{x \to \infty} \frac{1+\frac{3}{x}-\frac{12}{x^2}}{-6+\frac{1}{x}-\frac{5}{x^2}}=-\frac{1}{6}\\\\\\2)\; \; \lim\limits _{x \to 3}\frac{x^2-x-6}{x^2-6x+9}=\lim\limits _{x \to 3}\frac{(x-3)(x+2)}{(x-3)^2}=\lim\limits_{x \to 3}\frac{x+2}{x-3}=\infty \\\\\\3)\; \; \lim\limits_{x \to -3}\frac{4x^2+11x-3}{3x^2+10x+3}=\lim\limits_{x \to -3}\frac{4\, (x+3)(x-0,25)}{3\, (x+3)(x+\frac{1}{3})}=\lim\limits_{x \to -3}\frac{4x-1}{3x+1}=\frac{13}{8}$

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