That's your definition for continuity? Well I suggest looking the graph of $f$ and see if there are any "holes" or "jumps".
Let's do this in a rigorous manner:
The $\epsilon-\delta$ definition: Let $a\in X\subseteq \mathbb{R}$ and $f:X\to \mathbb{R}$. Then we say that $f$ is continuous at $a$ if \begin{equation}\forall \epsilon>0\exists \delta>0:\ \left|x-a\right|<\delta\implies \left|f(x)-f(a)\right|<\epsilon\end{equation} If $f$ is continuous at every point of $A\subseteq X$ then $f$ is said to be (pointwise) continuous on $A$.
It is usually hard to prove continuity using the $\epsilon-\delta$ definition. When one has studied limits of sequence this becomes much simpler using the following equivalent definition due to Heine :
Let $a\in X\subseteq \mathbb{R}$ and $f:X\to \mathbb{R}$. Then $f$ is continuous at $a$ iff for every sequence $(x_n)$ of elements of $X$, $x_n\to a\implies f(x_n)\to f(a)$
Using this we have to prove that if $x_n\to a$ for arbitrary $a\in \mathbb{R}$ then $f(x_n)\to f(a)$ or equivalently $2x_n^4-2\to 2a-2$. I will let you prove that.
Sidenote:
Even as is, your non rigorous definition is wrong because it allows essential discontinuities to be continuities. Look at the graph of $f(x)=\sin (1/x)$ $x\neq 0$ and $f(0)=0$. Is $f$ continuous?
EDIT: We need to prove that $x_n\to a\implies f(x_n)\to f(a)$ not for a particular $x_n,a$ but for all sequences $x_n$ and points $a$.
Now let $x_n$ be a sequence and $a\in \mathbb{R}$ so that $x_n\to a$. Then $f(x_n)=2x_n^4-2$ while $f(a)=2a^4-2$. We need to prove that $f(x_n)\to f(a)\iff 2x_n^4-2\to 2a^4-2\iff x_n^4\to a^4$ (the equivalencies are due to the limit laws). So the problem moves to proving that $x_n^4\to a^4$. But this also follows from the limit laws as $x_n\to a\implies x_n\cdot x_n\to a\cdot a\implies...\implies x_n^4\to a^4$