Chapter 2 - Practice Exercises - Page 111: 35

$f$ cannot be extended to be continuous at $x=1$ and $x=-1$.

$$f(x)=\frac{x(x^2-1)}{|x^2-1|}$$ We will need to draw the graph of the function $f$ in this exercise. The graph has been drawn and enclosed below. - Condition for continuous extension for function $f(x)$ at a point $c$: $\lim_{x\to c}f(x)$ must exist. Here, from the graph, we see that - As $x\to1^-$, $f$ approaches $-1$; while as $x\to1^+$, $f$ approaches $1$. So $\lim_{x\to1}f(x)$ does not exist, since $f$ does not approach any same number as $x\to1$ from both left and right. - Similarly, as $x\to-1^-$, $f$ approaches $-1$; while as $x\to-1^+$, $f$ approaches $1$. So $\lim_{x\to-1}f(x)$ does not exist, since $f$ does not approach any same number as $x\to-1$ from both left and right. Therefore, function $f$ cannot be extended to be continuous at $x=1$ and $x=-1$, as both limits there do not exist. The existence of limit is essential, because we need a value to extend the value of $f(c)$ to be equal with so that $f$ is continuous at $c$.