Question

lim стремится к бесконечности 7x^5+6x^4-x^3/2x^2+6x+1

Answer 1

Ско-о-о-бки

$\displaystyle \lim_{x \to \infty} \frac{7x^5+6x^4-x^3}{2x^2+6x+1}=\frac{\infty}{\infty}=\lim_{x \to \infty} \frac{x^5(7+\frac{6}{x}_{\to0}-\frac{1}{x^2}_{\to0})}{x^5(\frac{2}{x^3}_{\to0}+\frac{6}{x^4}_{\to0}+\frac{1}{x^5}_{\to0})}=\lim_{x \to \infty}\frac{7}{0}=\infty$