We define function $f(x) = 5x^3$.
To prove: $f$ is continuous in x, with $x \in \mathbb R.$
Suppose, $\epsilon > 0,$ we choose $\delta = \frac{\epsilon}{5(3x^2 + |3x| +1)}$, then for all $x \in \mathbb R$, with $|x - c| \lt \delta$, we have $|5x^3 - 5c^3| < \epsilon$. We also suppose $\delta < 1.$
|$5x^3 - 5c^3$| = 5|$x^3 - c^3$| = 5|$x - c$||$x^2 + c^2$ + cx| < 5 δ|$x^2 + c^2$ + cx|
|$x^2+c^2+xc$| < |$x^2 + (x + 1)^2+ x(x+1)$| so, |$x^2+c^2+xc$|< $3x^2 + |3x| + 1$
5 δ|$x^2 + c^2$ + cx| < 5 δ ($3x^2 + |3x| + 1$) < ϵ.
Is this proof sufficient?
No, you can't have your $\delta$ depending on $x$, as you haven't said anything about $x$ yet.
You'll need to 'work backwards' as I've always been taught, begin from what you want to prove to get an expression for $\delta$.