You are having trouble handling the denominator. If you've never seen the technique used to deal with that sort of thing it may be very mysterious and confusing to figure out how to proceed. So let me do this: I'll give an $\epsilon$-$N$ proof for:
Let $(s_n)$ be a sequence of real numbers. If $s_n\to s\neq 0$, then the sequence $\left(\frac{1}{s_n}\right)$ converges to $\frac{1}{s}$.
Let $\epsilon\gt 0$. The idea is that if $s_n$ is very close to $s$, then $\frac{1}{s_n}$ will be very close to $\frac{1}{s}$. The problem is that $\frac{1}{s_n}-\frac{1}{s} = \frac{s-s_n}{s_ns}.$ So while we can make the numerator small, the denominator is also changing, so that presents an obstacle to a simple chain of inequalities.
So: the first thing to do is to show that we can make sure that $s_ns$ is not too small; since we want to say that the expression is "less than or equal" to something, we need to be able to say $\left|\frac{1}{s_ns}\right|\leq k$ for some $k$. But in order to ensure that this quotient is small, we need to make sure that $s_ns$ is large. That is, we need to find some lower bound to $|s_ns|$.
Let $\ell = |s|\neq 0$. We know that there exists $N_1\gt 0$ such that if $n\geq N_1$, then $|s_n-s|\leq \frac{\ell}{2}$. Since $|s|-|s_n| \leq |s_n-s| \leq \frac{\ell}{2} = \frac{|s|}{2},$ then we have that if $n\geq N_1$, then $|s| - \frac{|s|}{2} \leq |s_n|,$ that is, $|s_n| \geq \frac{|s|}{2};$ and therefore $|s_n|\,|s| \geq \frac{|s|^2}{2},$ so if $n\geq N_1$, then $\frac{1}{|s_n|\,|s|} \leq \frac{2}{|s|^2}.$ So we have succeded in bounding above $\frac{1}{|s_ns|}$, by bounding $|s_ns|$ below ("away from $0$").
Now: let $\epsilon\gt 0$. We need to show that there exists an $N\gt 0$ such that for all $n\geq N$, $|\frac{1}{s_n} - \frac{1}{s}|\lt \epsilon$. What we will want to do is manipulate the latter inner product as: $\left|\frac{1}{s_n} - \frac{1}{s}\right| =\left|\frac{s-s_n}{s_ns}\right| = \frac{1}{|s_ns|}\left|s-s_n\right|.$ We know that if $n\geq N_1$, then $\frac{1}{|s_ns|}\leq \frac{2}{|s|^2}$. We also know that there exists $N_2\gt 0$ such that for all $n\geq N_2$, we have $|s-s_n| \lt \frac{|s|^2\epsilon}{2}.$ We know this because $s_n\to s$. So let $N=\max(N_1,N_2)$. Then if $n\geq N$, we will have $\text{both}\quad \frac{1}{|s_ns|}\leq \frac{2}{|s|^2}\qquad\text{and}\qquad |s-s_n|\lt \frac{|s|^2\epsilon}{2}.$ Therefore, if $n\geq N$, then $\begin{align*} \left|\frac{1}{s_n} - \frac{1}{s}\right| &= \left|\frac{s-s_n}{s_ns}\right| \\ &= \frac{|s-s_n|}{|s_ns|}\\ &= \frac{1}{|s_ns|}|s-s_n|\\ &\leq \left(\frac{2}{|s|^2}\right)|s-s_n|\\ &\lt \left(\frac{2}{|s|^2}\right)\left(\frac{|s|^2\epsilon}{2}\right)\\ &= \epsilon, \end{align*}$ and so we have shown that $\frac{1}{s_n}\to \frac{1}{s}$.
Now, think about what you have here. Your problem is that you have a denominator that depends on $s_n$ and $s$; so you want to bound that denominator "away from $0$"; this will give you control over the denominator. Then you can also bound $|s_n^3 - s^3|$, which will give you control over the numerator. Having control over both will give the desired result.