Question

limx→∞ ^3√(x^3−4x^2−7x−5√3/x^2−7x+9

Answer 1

is this your problem \(\displaystyle\lim_{x\to \infty}\frac{\sqrt[3]{x^3-4x^2-7x-5\sqrt{3}}}{x^2-7x+9}\) ?

Answer 2

yes thats the one!

Answer 3

0

Answer 4

can u explane how did u get it?

Answer 5

the higest power or x in the numerator is essentually just x (the cube root of x^3...the other terms in the numerator really don't matter)

and on the bottom is x^2

power on the bottom is larger than the top so the limit is 0

Answer 6

thanx!