Ответ:

[tex]\displaystyle +\infty[/tex]

Объяснение:

[tex]\displaystyle \lim_{x\to \infty}\frac{x^4-5x}{x^2-3x+1}=\\\\\lim_{x\to \infty}\frac{x^4(\frac{x^4}{x^4}-\frac{5x}{x^4})}{x^2(\frac{x^2}{x^2}-\frac{3x}{x^2}+\frac{1}{x^2})}=\\\\\lim_{x\to \infty} \frac{x^2(1-\frac{5}{x^3})}{1-\frac{3}{x}+\frac{1}{x^2}}=\\\\\lim_{x\to \infty} \frac{x^2}{1}=+\infty[/tex]