- Continuous everywhere.
- Continuous at integer points only.
- Continuous at non-integer points only.
- Differentiable everywhere.
Solution:
f(x) = x - x
Consider n be an integer.
$\text{f(x)}=\text{x}-[\text{x}]=\begin{cases}\text{x}-(\text{n}-1)&\text{n}-1\leq\text{x}<\text{n}\\0&\text{x}=\text{n}\\\text{x}-\text{n}&\text{n}\leq\text{x}<\text{n}+1\end{cases}$
Now,
LHL at x = n
$=\lim\limits_{\text{x}\rightarrow\text{n}^{-}}\text{f(x)}=\text{x}-\text{n}-1=\text{x}-\text{n}+1$
RHL at x = n
$=\lim\limits_{\text{x}\rightarrow\text{n}^{+}}\text{f(x)}=\text{x}-\text{n}=\text{x}-\text{nAs},$
$\text{LHL}\neq\text{RHL}$ at x = n
i.e., given function is not continuous at n.
Now, n is any integer. Therefore, given function is not continuous at integers.
Therefore, given points are continuous at non-integer points only.
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