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

Dec 25, 2016

${\lim}_{t \to \infty} \left(\sqrt[3]{{t}^{3} + 1} - t\right) = 0$

#### Explanation:

Use the difference of cubes identity:

${a}^{3} - {b}^{3} = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$

with $a = \sqrt[3]{{t}^{3} + 1}$ and $b = t$ as follows:

$\sqrt[3]{{t}^{3} + 1} - t = \frac{{\left(\sqrt[3]{{t}^{3} + 1}\right)}^{3} - {t}^{3}}{{\left(\sqrt[3]{{t}^{3} + 1}\right)}^{2} + t \left(\sqrt[3]{{t}^{3} + 1}\right) + {t}^{2}}$

$\textcolor{w h i t e}{\sqrt[3]{{t}^{3} + 1} - t} = \frac{\left({t}^{3} + 1\right) - {t}^{3}}{{\left(\sqrt[3]{{t}^{3} + 1}\right)}^{2} + t \left(\sqrt[3]{{t}^{3} + 1}\right) + {t}^{2}}$

$\textcolor{w h i t e}{\sqrt[3]{{t}^{3} + 1} - t} = \frac{1}{{\left(\sqrt[3]{{t}^{3} + 1}\right)}^{2} + t \left(\sqrt[3]{{t}^{3} + 1}\right) + {t}^{2}}$

Note that for positive values of $t$, all terms in the denominator are positive so:

So, when $t > 0$:

$0 < \frac{1}{{\left(\sqrt[3]{{t}^{3} + 1}\right)}^{2} + t \left(\sqrt[3]{{t}^{3} + 1}\right) + {t}^{2}} < \frac{1}{t} ^ 2$

So:

$0 \le {\lim}_{t \to \infty} \frac{1}{{\left(\sqrt[3]{{t}^{3} + 1}\right)}^{2} + t \left(\sqrt[3]{{t}^{3} + 1}\right) + {t}^{2}} \le {\lim}_{t \to \infty} \frac{1}{t} ^ 2 = 0$

So:

${\lim}_{t \to \infty} \left(\sqrt[3]{{t}^{3} + 1} - t\right) = {\lim}_{t \to \infty} \frac{1}{{\left(\sqrt[3]{{t}^{3} + 1}\right)}^{2} + t \left(\sqrt[3]{{t}^{3} + 1}\right) + {t}^{2}} = 0$