0

Correct me if I'm wrong, but for the following function, at x = $\sqrt2$

$$f(x) = \lfloor x^2-2\rfloor$$

it is discontinuous because the left hand limit $= -1$, while the right hand limit $= 0$. Hence, the limit of $f(x)$ does not exist and hence it is discontinuous.

As to the type of discontinuity, is it a jump discontinuity? Or a non-removable discontinuity

Alessio K
  • 10,599
xnajasho
  • 1
  • 1
    Jump-discontinuities are non-removable discontinuities... So to the answer of which is it? The answer is yes. – JMoravitz Sep 15 '20 at 17:17
  • Oh I did not learn that jump discontinuities are non-removable discontinuities. Thanks for the clarification! – xnajasho Sep 15 '20 at 17:26
  • 2
    The only removable type of discontinuity is one where the left and right limits both agree but the value at that point does not match the limit. That is, you have a removable discontinuity at a point $c$ for a function $f$ iff $\lim\limits_{x\to c^+}f(x)=\lim\limits_{x\to c^-}f(x)=\lim\limits_{x\to c}f(x) \neq f(c)$. It is "removable" in the sense that if you were to replace the value of $f(c)$ with the value of the limit instead then it would have been continuous at that point. All other discontinuities are non-removable. – JMoravitz Sep 15 '20 at 17:33
  • Just a slight precision on JMoravitz comment, $f(c)$ may not be defined either (e.g. $\frac{\sin x}{x}$ in $0$), still removable. – zwim Sep 15 '20 at 18:30

0 Answers0