How do you find #lim root3(t^3+1)-t# as #t->oo#?

1 Answer
Dec 25, 2016

#lim_(t->oo) (root(3)(t^3+1)-t) = 0#

Explanation:

Use the difference of cubes identity:

#a^3-b^3=(a-b)(a^2+ab+b^2)#

with #a=root(3)(t^3+1)# and #b=t# as follows:

#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 #t#, all terms in the denominator are positive so:

So, when #t > 0#:

#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#