0

I'm trying to find the continuity of the function $f(x) = \lfloor x^2 \rfloor$. I need to check if function $f$ is continuous at $0$.

It's in between $-1$ and $1$, since $f(-1) = 1$ and $f(1) = 1$ and it's not in $[0,1)$, since $f(0) = 0$ and $f(1) = 1$.

But the above observations is not correct according to Intermediate value theorem, since the first boundary $[-1,1]$ have the same result value. I'm confused here.

Ian
  • 101,645
Mitty
  • 217
  • The intermediate value theorem doesn't say anything when $f(a)=f(b)$. Instead, you should notice that if $x \in (-1/2,1/2)$ then $f(x)=0$, so $f$ is locally constant at zero, hence continuous there. – Ian Sep 16 '15 at 19:10
  • That's clear. I think they removed the floor function when edit was allowed. If there is a f(x) with floor(x^2) then the soln (-1/2,1/2) would be -1 and 0 right? – Mitty Sep 16 '15 at 19:15
  • If $x \in (-1/2,1/2)$ then $x^2 \in (0,1/4)$ and therefore $\lfloor x^2 \rfloor = 0$. – Ian Sep 16 '15 at 19:36

1 Answers1

0

put $‎\delta‎‎‎= ‎\sqrt{\frac{1}{3}}‎ $ then $-\sqrt{\frac{1}{3}}\leq x-0\leq \sqrt{\frac{1}{3}}$ so $0\leq x^2 \leq \frac{1}{3}‎\Longrightarrow‎ f(x) = [x^2]=0$

R.N
  • 4,318