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 ).

$\frac{x}{e^x- 1}$ is NOT continuous.  In order for a function, f(x), to be continuous at x= a, three things must be true.

*

*$\lim_{x\to a} f(x)$ exist.

*f(a) exist.

*$\lim_{x\to a} f(x)= f(a)$.
(For 3 to make sense, 1 and 2 must be true so often only
3 is stated.)

$\lim_{x\to 0} \frac{x}{e^x- 1}$ exists and is 0 (see below) so 1 is true.  But since $e^0- 1= 0$ and we cannot divide by 0, the function is not defined at 0. 2 is NOT true.  In order to have a function that is continuous at x= 0, we need to redefine it at x= 0 to be that limit, 1.
The MacLaurin series (Taylor series at 0) for e^x is
$1+ x+ x^2/2+ x^3/3!+ \cdot\cdot\cdot+ x^n/n!+ \cdot\cdot\cdot$.
$e^x- 1= x+ x^2/2+ x^3/3!+ \cdot\cdot\cdot+ x^n/n!+ \cdot\cdot\cdot$ so $\frac{x}{e^x- 1}= \frac{x}{x+ x^2/2+ x^3/3!+ \cdot\cdot\cdot+ x^n/n!+ \cdot\cdot\cdot}= \frac{x}{x(1+ x/2+ x^2/3!+ \cdot\cdot\cdot+ x^{n-1}/n!+ \cdot\cdot\cdot)}= \frac{1}{1+ x/2+ x^2/3!+ \cdot\cdot\cdot+ x^{n-1}/n!+ \cdot\cdot\cdot}$
The limit as x goes to 0 is $\frac{1}{1}= 1$.
