Prove $$f(x)=\frac{e^x-1}{x}$$ is (dis)continuous at $x=0$.
$$\mathcal D_f=\mathbb R\setminus\{0\}$$ Let: $f=\overset{\sim}{f}_{\mid\mathbb R\setminus\{0\}}$ and $$\overset{\sim}{f}:\mathbb R\to\mathbb R$$ given by the same formula and $$\overset{\sim}{f}(0):=\lim_{x\to 0} f(x)$$ By the theorem:
Function $f:I\to\mathbb R$ is continuous at $x=c\in I$ iff it has a limit at the point $c\in I$ and it equals $f(c)$.
$\overset{\sim}f$ satisfies that condition.
Is this sufficient?
edit: Whatever the person who typed the question I got might have been thinking, I'll take into account the following, as stated in the comments as well:
A function $f$ is continuous at $c$ if the following three conditions are met for continuity at a point
$(1)$ $f(c)$ is defined.
$(2)$ $\exists\displaystyle\lim_{x\to c}f(x)$ (real limit, not infinite)
$(3)$ $\displaystyle\lim_{x\to c}f(x)=f(c)$
In the case of $\overset{\sim}f$ possible discontinuity could be classified as removable.
The function should be $$ f(x)=\begin{cases} \frac{e^x-1}{x} & x\neq 0\\ 1&x=0. \end{cases}$$
To prove continuity, you need to show that $\lim_{x\to 0^-}f(x) = \lim_{x\to 0^+}f(x)=f(0)=1$. Can you complete it now?