I Understanding Removable discontinuity of the given function: ##e^x##

[pbuk]

If you focus only on what the rigorous maths is telling you, then $1/x$ is a continuous function. There is no other conclusion.

I am not disputing that. I am not even disputing that it is possible to define a discontinuity at $x$ in such a way that $x$ must be in the domain of $f$.

All I am saying is that if you remove that requirement from the definition you have something that can be even more useful and does not lead to any mathematical inconsistency. To illustrate its utility, if I say "$x^2 / x$ has a removable discontinuity at $x = 0$" then surely everyone understands what I mean.

I don't think it is better to have to say "$f: \mathbb R | 0 \to \mathbb R; x \mapsto \frac{x^2}x$ does not have a removable discontinuity at $x = 0$ because 0 is not in the domain of $f$, however $$ g: \mathbb R \to \mathbb R; \\
x \mapsto \begin{cases}
\frac{x^2}x & x \ne 0 \\
0 & x = 0
\end{cases}
$$(noting that $g(x) = f(x)$ for all $x$ in the domain of $f$) does have a removable discontinuity at $x = 0$".

 

[PeroK]

All I am saying is that if you remove that requirement from the definition you have something that can be even more useful and does not lead to any mathematical inconsistency. To illustrate its utility, if I say "$x^2 / x$ has a removable discontinuity at $x = 0$" then surely everyone understands what I mean.

Yes, but technically that function is a continuous function (on its domain). If you look at the definition of "removable discontinuity" and "continuity" you see that a continuous function, technically, may have a removable discontinuity. The contradiction is only linguistic.

I don't think it is better to have to say "$f: \mathbb R | 0 \to \mathbb R; x \mapsto \frac{x^2}x$ does not have a removable discontinuity at $x = 0$ because 0 is not in the domain of $f$, however $$ g: \mathbb R \to \mathbb R; \\
x \mapsto \begin{cases}
\frac{x^2}x & x \ne 0 \\
0 & x = 0
\end{cases}
$$(noting that $g(x) = f(x)$ for all $x$ in the domain of $f$) does have a removable discontinuity at $x = 0$".

That function does have a removable discontinuity at $x = 0$. But, that does not imply it is not a continuous function. That's the point I made in post #4:

Note that technically a function has to be defined at a point in order to be discontinuous. So, technically, the function $1/x$ is not discontinuous at $x =0$. That said, it still tends to be called a discontinuity.

 

[pbuk]

I agree with everything you say in #32, particularly that we may speak of a removable discontinuity of a continuous function and

The contradiction is only linguistic.

 

[Svein]

In complex analysis the definition of a removable singularity goes like this: Suppose that f(z) is analytic in the region Ω', obtained by omitting a point a from a region Ω. A necessary and sufficient conditiion that there exist an analytic function in Ω which coincides with f(z) in Ω' is that \lim_{ z\to a}(z-a)f(z)=0. The extended function is uniquely defined.
Nor do we have a problem with \frac{1}{z}. Cauchy's formula says that f(z)=\frac{1}{2\pi i}\int_{C}\frac{f(\zeta) d\zeta}{\zeta - z} for all f(z) that are analytic inside of the closed curve C.

 

[Office_Shredder]

@Svein I'm not sure how that addresses the question of whether $1/z$ is a continuous function though, which is the analogous question to $1/x$ in $\mathbb{R}$

 

[Svein]

@Svein I'm not sure how that addresses the question of whether $1/z$ is a continuous function though, which is the analogous question to $1/x$ in $\mathbb{R}$

I did not say that $1/z$ was continuous. I said that in complex analysis we do not have a problem with $1/z$. It is called a pole.