epsilon delta proof that limit $x+t^3$ as t approaches a is $x+a^3$

Question

Found a small exercise on page 101 of Spivak's Calculus. I'm having some trouble with a clean proof that works even when $a = 0$ that doesn't rely on limit theorems. My naive approach: $$|{f(t)-l}| = |{(x-t^3)-(x-a^3)}| = |{t^3-a^3}| = |{t-a}|*|{a^2+at+t^2}|$$. Then, if $|t-a| < \frac{\epsilon}{|a^2+at+t^2|}, |t^3-a^3| < \epsilon$, but this only works as long as the denominator is not 0. Any hints? My first idea is to bound $|t^3-a^3|$ above by a polynomial that is always $\geq 0$ but which?

Answer 1

If $|t-a|<\delta$ then $|t|<|a|+\delta.$ Hence $$|a^2+at+t^2|<a^2+a^2+|a|\delta+a^2+\delta^2+2|a|\delta=3a^2+3|a|\delta+\delta^2.$$

Now fix $\epsilon$ and choose a $\delta$ such that $$\delta(3a^2+3|a|\delta+\delta^2)<\epsilon.$$ (This is of course always possible, do you see why ?). For this $\delta$, you get $$|f(t)-l|<\epsilon.$$