Question

#6 помогите. (lim)
Хотя бы 3 решите

Answer 1

$\lim\limits _{x \to \infty}(\frac{7}{x^3}+\frac{6}{x^2})=0+0=0\\\\\lim\limits _{x \to \infty} \Big ((\frac{1}{x^2}+2)\cdot (\frac{3}{x^3}-4)\Big )=2\cdot (-4)=-8\\\\\lim\limits _{x \to \infty}\frac{1+2x}{3x-7}=\lim\limits _{x \to \infty}\frac{\frac{1}{x}+2}{3-\frac{7}{x}}=\frac{2}{3}\\\\\lim\limits _{x \to \infty}(x^2+x-2)=\infty$