Question

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

Answer 1

`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`