0

The possible points are $0, -1, 1$

Obviously, $\mathrm{sign}(x^3-x)$ is discontinuous at $x=0$.

The function is supposed to be discontinuous at $\pm 1$ and continuous at $0$

Checking continuity at $x=1$

$$\lim_{x\to 1^+} f(x)$$ $$=\lim_{h\to 0} |1+h| \mathrm{sign}((1+h)^3-(1+h))$$ $$=0$$

And the left hand limit is also $0$

How is the function discontinuous at $\pm 1$? I feel that I making a computation error, but the signum function is throwing me off

Vincenzo Tibullo
  • 10,955
Aditya
  • 6,191

1 Answers1

1

Hint:

What are the signs of $x^3-x$ below $-1$, between $-1$ and $0$, between $0$ and $+1$, and above $+1$?

What are the values of $f(x)$ at $-1-\delta,-1+\delta,0-\delta,0+\delta,+1-\delta,+1+\delta$ when $0 \lt\delta \ll 1$, and so the left and right limits as $\delta \to 0^+$?

  • $f(-1-\delta)=+1+\delta \to +1$
  • $f(-1+\delta)=-1+\delta\to -1$
  • $f(0-\delta)=0-\delta\to 0$
  • $f(0+\delta)=0-\delta\to 0$
  • $f(+1-\delta)=-1+\delta\to -1$
  • $f(+1+\delta)=+1+\delta\to +1$
Henry
  • 157,058