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Solution

Let $c$ be an integer. For LHL, $ \lim_{x \to c^-} f(x)=\lim_{h\to0^+}[(c-h)-[c-h]] $ Since $[c-h]=c-1$, $ =\lim_{h\to0}(c-h-(c-1)) $ $ =\lim_{h\to0}(1-h)=1 $ For RHL, $ \lim_{x \to c^+} f(x)=\lim_{h\to0^+}[(c+h)-[c+h]] $ Since $[c+h]=c$, $ =\lim_{h\to0}(c+h-c) $ $ =\lim_{h\to0}h=0 $ Thus, $ \text{LHL} \ne \text{RHL} $ $\therefore f(x)$ is discontinuous at all integers. Hence, $f(x)$ is continuous at non-integer points only.