Largest subset on which a function is continuous Let $f: \mathbb{C} \to \mathbb{C}$ a function with  $$f(x) =0, ~~~ \text{if} ~~ x = 0 $$ and 
$$f(x) = (e^x - 1)/x, ~~~\text{if} ~~x \neq 0$$ 
I want to determine the largest subset $A \subset \mathbb{C}$ in which $\phi$ is continuous. I think this is $A = \mathbb{C}\backslash\{0\}$, but so far I did not succeed in proving this with the epsilon and delta definitions. Any suggestions?
EDIT: I could use that the sum and the product of continuous functions on $\mathbb{R}\backslash\{0\}$ are also continuous. But is there a more 'formal' way?
 A: Let me write $z=x+iy$ instead of simply $x$, since usually $z$ represents a complex number while $x$ represents its real part. Let's look at $f(z)=\frac{e^z-1}{z}$ when $z\neq0$. Both $h(z)=e^z-1$ and $g(z)=z$ are continous functions in $\Bbb{C\setminus\{0\}}$, and because $z\neq0$,$\ $ $h(z)\neq0$ for all $z\in \Bbb{C\setminus\{0\}}$, so $f(z)=\frac{h(z)}{g(z)}$ is continous on $\Bbb{C\setminus\{0\}}$.
Now, to check whether $f$ is continous at $0$, we need to prove or disprove that
$$\lim_{z\to 0}f(z)=f(0)=0$$
In order to help us in this case, the following lemma is useful:

Lemma. Let $f, g\in H(D(z_o;r))$ with $r>0$. Suppose that $$\lim_{z\to z_0}f(z)=\lim_{z\to z_0}g(z)=0$$
  and
  $$\lim_{z\to z_0}\frac{f'(z)}{g'(z)}=\lambda$$
  Then
  $$\lim_{z\to z_0}\frac{f(z)}{g(z)}=\lambda$$

Proof. 
$$\lim_{z\to z_0}\frac{f(z)}{g(z)}=\lim_{z\to z_0}\frac{f(z)-f(z_0)}{z-z_0}\cdot\frac{z-z_0}{g(z)-g(z_0)}=\frac{f'(z_0)}{g'(z_0)}=\lambda.$$
The functions $h(z)$ and $g(z)$ defined before satisfy the Lemma's hypothesis, so
$$\lim_{z\to 0}f(z)=\lim_{z\to 0}\frac{h(z)}{g(z)}=\lim_{z\to 0}\frac{h'(z)}{g'(z)}=\lim_{z\to 0}\frac{e^z}{1}=e^0=1$$
But $$1=\lim_{z\to z_0}f(z)\neq f(0)=0$$
So $f$ is not continnous at $z=0$, and therefore, the largest subset of $\Bbb{C}$ on which it is continous is $\Bbb{C}\setminus\{0\}$.
