# Question on the task: Prove $f(x)=\frac{e^x-1}{x}$ is (dis)continuous at $x=0$.

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.

• $f(x)$ is not defined at $x=0$, so how could it be continuous there? Feb 17, 2020 at 19:14
• @J.W.Tanner, I was confused by the question, but I thought our professor expects us to expand the function, however, only the expansion is continuous at $x=0$ so I posted it here just in case. Feb 17, 2020 at 19:20
• Yes the function works but the point of the excercise is to show that the limit exists. Feb 17, 2020 at 19:28
• @ms._VerkhovtsevaKatya: the answer is circular. If you complete the function with its limit at $x=0$, you implicitly choose a continuous extension. And the new function is perforce continuous.
– user65203
Feb 17, 2020 at 19:57

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?