The value of undersetx→1lim(4x^3-x^2+2x-5)/(x^6+5x^3-2x-4) is:

$\underset{x→1}{\lim}\frac{4x^3-x^2+2x-5}{x^6+5x^3-2x-4}$$[\frac{0}{0}form]$

On applying L’Hospital’s rule, we get :

$\underset{x→1}{\lim}\frac{12x^2-2x+2}{6x^5+15x^2-2}=\frac{12-2+2}{6+15-2}=\frac{12}{19}$