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.

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    $\begingroup$ $f(x)$ is not defined at $x=0$, so how could it be continuous there? $\endgroup$ Feb 17, 2020 at 19:14
  • $\begingroup$ @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. $\endgroup$
    – Invisible
    Feb 17, 2020 at 19:20
  • $\begingroup$ Yes the function works but the point of the excercise is to show that the limit exists. $\endgroup$ Feb 17, 2020 at 19:28
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    $\begingroup$ @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. $\endgroup$
    – user65203
    Feb 17, 2020 at 19:57

1 Answer 1


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?


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