$\lim \limits_{x \to -\infty} (1-5x)^{1/3}=\infty$
Therefore we need to find $\forall M>0$ $\exists \delta=\delta(M,x_0)$ such that $f(x)>M $ and $0<|x-x_0|<\delta$
So $(1-5x)^{1/3}>M$
$1-5x>M^3$
$x>\frac{1-M^3}{5}$
so set $\delta=max${$\frac{1-M^3}{5}$}?
No, limit at infinity works a little differently than limit at a point. You don't have any $x_0$ to work with, so some of the things you wrote don't make sense.
Intuition: when showing $f(x) \to \infty$ as $x \to x_0$, you need to prove that for any $M$, $f(x) > M$ when you get "close enough" to $x_0$. "Close enough" to $x_0$ means finding some $\delta(M)$ such that etc.
When showing the same as $x \to -\infty$, you need to prove that for any $M$, $f(x) > M$ when you get "close enough" to $-\infty$. "Close enough" to $-\infty$ means finding some $a(M)$ s.t. if $x < a(M)$, then $f(x) > M$.
Your calculations were correct until $1-5x>M^3$, but then you made a sign mistake. Re-check your work and find a suitable $a(M)$ so that for all $x$ less than it, $f(X) > M$.