Answer
No, see explanations.
Work Step by Step
Step 1. For $x\lt-1, x\ne-1$, $f(x)=\frac{x(x^2-1)}{|x^2-1|}==\frac{x(x^2-1)}{x^2-1}=x$, $\lim_{x\to-1^-}f(x)=-1$
Step 2. For $-1\lt x\lt1, x\ne-1, x\ne1$, $f(x)=\frac{x(x^2-1)}{|x^2-1|}==-\frac{x(x^2-1)}{x^2-1}=-x$, $\lim_{x\to-1^+}f(x)=-(-1)=1$, and $\lim_{x\to1^-}f(x)=-(1)=-1$
Step 3. For $x\gt1, x\ne1$, $f(x)=\frac{x(x^2-1)}{|x^2-1|}==\frac{x(x^2-1)}{x^2-1}=x$, $\lim_{x\to1^+}f(x)=1$
Step 4. Graph the function as shown.
Step 5. We can see that the continuity can not be extended at $x=1$ or $x=-1$ because the limits do not exist at these points.
