As w, x $\in$ { 0 + 1 }*.
By taking any string $\in$ $\sum$* says S examples.
S = 10001010101
We can say that here w = $\in$, so $w^{r}$ is also $\in$. And x = 10001010101.
So any string $\in$ $\sum$* can be accepted like above argument.
Therefore language generated will be $\sum$*, which is a regular language.
Interesting question would be
L = {w$w^{r}$x ∣ x,w $\in$ {0,1}^+}
Note here x, w $\in$ {0 + 1}^+.
Here we can't say w = empty string. It should have at least length greater than or equal to 1.
we know that w$w^{r}$ can be accepted by an NPDA (It is a very popular example). NPDA because we can't determine the middle of the string.
Now it is easy to see that w$w^{r}$x can also be accepted by NPDA by some modification. When we are done with w$w^{r}$ part and our stack consisting z(Stack symbol) then if we see either 0 or 1 we can move to the final state.
For More such examples.
http://gatecse.in/identify-the-class-of-a-given-language/
