How do you find #lim root3(t^3+1)-t# as #t->oo#?
1 Answer
Explanation:
Use the difference of cubes identity:
#a^3-b^3=(a-b)(a^2+ab+b^2)#
with
#root(3)(t^3+1)-t = ((root(3)(t^3+1))^3-t^3)/((root(3)(t^3+1))^2+t(root(3)(t^3+1))+t^2)#
#color(white)(root(3)(t^3+1)-t) = ((t^3+1)-t^3)/((root(3)(t^3+1))^2+t(root(3)(t^3+1))+t^2)#
#color(white)(root(3)(t^3+1)-t) = 1/((root(3)(t^3+1))^2+t(root(3)(t^3+1))+t^2)#
Note that for positive values of
So, when
#0 < 1/((root(3)(t^3+1))^2+t(root(3)(t^3+1))+t^2) < 1/t^2#
So:
#0 <= lim_(t->oo) 1/((root(3)(t^3+1))^2+t(root(3)(t^3+1))+t^2) <= lim_(t->oo) 1/t^2 = 0#
So:
#lim_(t->oo) (root(3)(t^3+1)-t) = lim_(t->oo) 1/((root(3)(t^3+1))^2+t(root(3)(t^3+1))+t^2) = 0#