Домашнее задание (1,2)

Решите задачу:

$\lim_{x\to \infty}\frac{3x^5+2x^7}{2x^5-8x^7}=[delim\; na\; x^7]=\lim_{x\to \infty}\frac{\frac{3}{x^2}+2}{\frac{2}{x^2}-8}=\\\\=[\frac{3}{x^2}\to 0,\frac{2}{x^2}\to 0]=-\frac{2}{8}=-\frac{1}{4}\\\\\\\lim_{x\to \infty}\frac{4x^5-3x+2}{3+2x^3+5x^5}=[delim\; na\; x^5]=\lim_{x\to \infty}\frac{4-\frac{3}{x^4}+\frac{2}{x^5}}{\frac{3}{x^5}+\frac{2}{x^2}+5}=\frac{4}{5}$