Is the function continuous at x=0? Check if the function $f$ is continuous.
$f(x)=$\begin{matrix}
0 & ,x=0\\ 
\frac{1}{[\frac{1}{x}]} & ,0<x\leq 1 
\end{matrix}.
For $0<x\leq 1$,,f is continuous because it is fraction of continuous functions.
How can I check if it is continuous at $x=0$?

$$|f(x) - f(0)| =  |\frac 1 {[\frac 1 x]} - 0| = |\frac 1 {[\frac 1 x]}| = \frac 1 {[\frac 1 x]} \le \frac 1 {\frac 1 x} =  x  = x - 0\;\; (\text {since $x > 0$ and $\frac 1 {[\frac 1 x]} \le \frac 1 {\frac 1 x}$})$$ 
Given any $\epsilon \gt 0 $ however small, $\exists \ \delta( = \epsilon) \gt 0 $ such that $|f(x) - f(0)| \lt \epsilon$ whenever $x = x - 0 \lt \delta$
$\implies \lim_{x \rightarrow 0^+} f(x) = f(0)\implies f$ is right-continuous at $x = 0$.
