Prove with epsilon delta the limit of $5x^3$

Question

I try to prove the $\lim_{x \to a}5x^3$ with the epsilon-delta theorem for every real a.

I already came up with the idea of $0<|x-a|<\delta$

Since $|5x^3 - 5a^3| = 5|x-a||(x-a)^2 +3ax|$

Let $\delta<1$ and so $-1<x-a<1$

However, I don't know how to find out what $3ax$ is? Can someone please help me??

Answer 1

Hint: Use the identity $$ x^3 - y^3 = (x-y)(x^2 + xy + y^2) $$

you then have $$ |x^3 - a^3| < \epsilon \\ \Leftarrow |x-a| (x^2 + |xa| + a^2) < \epsilon \\ \Leftarrow |x-a| < \epsilon / R\ \ \ \& \ \ \ (x^2 + |xa| + a^2) < R $$

Now you can prove that if $|x-a|<\delta$ then $(x^2 + |xa| + a^2) $ is bounded by a certain $R(\delta)$.