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 ).
 A: Try and use the definition of continuity and the fact that $\displaystyle \lim_{x\to 0}\frac{x}{e^{x}-1} = 1$ .
So a perfectly okay argument would be that since $\displaystyle \lim_{x\to 0} f(x)= \lim_{x\to 0}\frac{x}{e^{x}-1} = 1=f(0)$ we have $f$ is continuous at $0$ and since it is a product of continuous functions for $x\neq 0$ , you have that $f$ is continuous.
If you want a full rigourous argument then try and show that for any $\epsilon>0$ there exists a $\delta>0$ such that $|x|<\delta \implies |\frac{x}{e^{x}-1}-1|<\epsilon $ .
To check that $f'(0)$ exists, you need to show that $$\lim_{h\to 0} \frac{f(h)-f(0)}{h-0}=\lim_{h\to 0} \dfrac{\frac{h}{e^{h}-1}-1}{h}$$
exists . That can be done in several ways. The most useful way would be to consider the Taylor expansion of $e^{h}$ and work them out. Try to show that $f'(0)=\frac{-1}{2}$ .
Spoiler
$$h-e^{h}+1= -\frac{h^{2}}{2!} - \frac{h^{3}}{3!}-...$$
And $h(e^{h}-1)=h^{2}+\frac{h^{3}}{2!}+...$
So you divide them out and see that you get $\displaystyle\lim_{h\to 0}\frac{-\frac{h^{2}}{2}+o(h^{2})}{h^{2}+o(h^{2})}=\frac{-1}{2}$
A: $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 $x=0$
A: $\frac{x}{e^x- 1}$ 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 $\frac{1}{1}= 1$.
