Question

lim x->к бесконечности 5х^4-х^3+2х/х^4-8х^3+1

Answer 1

$lim_{xtoinfty}frac{5x^4-x^3+2x}{x^4-8x^3+1}=lim_{xtoinfty}frac{x^4(5-frac{1}{x}+frac{2}{x^2})}{x^4(1-frac{8}{x}+frac{1}{x^4})}=lim_{xtoinfty}frac{5-frac{1}{x}+frac{2}{x^2}}{1-frac{8}{x}+frac{1}{x^4}}= \ =frac{5-0+0}{1-0+0}=frac{5}{1}}=5$