Why is $f(x)=\frac{x}{e^x-1}$ continuous? My exercise says:
$$f(x)=\frac{x}{e^x-1},\quad x\neq0$$
$$f(0)=1$$
Can someone explain whatever this means? And why does the graph not cut off at $x=0$?
Edit : Apologies ,it's not $f(0)=0$ but $f(0)=1$.
And to clarify the exercise asks to confirm that $f'(0)$ exists (But that's not what puzzles me right now ).

$f(x)=$$ \begin{cases} 
      0 & x= 0 \\
      f(x)=\frac{x}{e^x-1} &x\neq0 \\
   \end{cases}
$
in order for the function to be continous at a point then this condition must be satsified which is
$\lim_{x \to a} f(x)=f(a)$
but $\lim_{x \to 0} \frac{x}{e^x-1}=1$ which is not equal to $f(0)$ so  then $f$ is discontinuous at $x=0$
