we need to show it is discontinuous at x≠0 can any one just explain to me to me answer

Q)
   $$f(x)=\begin{cases}
x &\text{if }x\in \mathbb{Q} \\
0 &\text{if }x\in \mathbb{R}\setminus\mathbb{Q}
\end{cases}$$
  we need to show it is discontinuous at x≠0 
Answer
  Let us test the discontinuity at x=c, where c in Q and $c≠0$.
  $f(c+)=\lim f(x)=\lim f(c+h)-f(c)\backslash h$ = $$f(x)=\begin{cases}
\lim ((c+h)-(c))\backslash h &\text{if }c+h\in \mathbb{Q} \\
\lim ((0)-(c))\backslash h,doesnot exit  &\text{if }c+h\in \mathbb{R}\setminus\mathbb{Q}
\end{cases}$$
  where is 
  $x\rightarrow c+$

$h\rightarrow 0$
my question
this is the definition right of right side limit , please tell me this definition right or wrong?
what is h ?
 A: Hint: Use the result, a function is continuous if for any sequence $(x_n)_{n\in\mathbb{N}}$ of points in the domain which converges to $c$, the corresponding sequence $\left(f(x_n)\right)_{n\in \mathbb{N}}$ converges to $f(c)$. In other words, $$\forall (x_n)_{n\in\mathbb{N}} \subset I:\lim_{n\to\infty} x_n=c \Rightarrow \lim_{n\to\infty} f(x_n)=f(c).$$
A: If function $f(x) $ is not continuous
at $x = a$ then we can construct a sequence $(x_n)$ which converges to $a$ but for which
$f(x_n)$ does not converge to f(a).
Take two sequences $(x_n)$ and $(y_n)$ of rationals and irrationals, respectively  and think about their images.
A: This is the so called Riemann function. If $x\not=0$ and $x\in \mathbb{Q}$, then $f(x)=x$. But if $f(x)$ is continuous then we can find a small neighborhood around $x$ such that for any $y$ in the neighborhood, $|f(x)-f(y)|$ is arbitrally small. This is impossible if $y$ is irrational. Similarly if $x$ is irrational, you may select a rational number $y$ because $\mathbb{Q}$ is dense in $\mathbb{R}$. This suggests $f(x)$ cannot be continuous except at $0$, at that point it is continuous. This is an exercise for you. 
A: The usual definition of "$f$ is continuous at $x = x_0$" is "$f(x_0) = \lim_{x \to x_0} f(x)$".  Here, $\lim_{x \to x_0} f(x)$ does not exist for any nonzero $x_0$, rational or not.  So by the usual definition of continuity, $f$ is not continuous at $x_0$.  Your question seems to involve the definition of the derivative, which you don't need here.
