Why is a function with a hole not considered to be continuous by the $\varepsilon-\delta$ definition?

Question

Say a function, $f(x)$, has a hole at $x=c$. So $f(c)$ does not exist. But say that the limit of $f(x)$ as $x$ approaches $c$ exists and is $L$. We can pick any positive number, $\varepsilon$, so that shifting $x$ sufficiently close to $c$ on the $x$-axis (so that the distance between $x$ and $c$ is always less than some number, $\delta$) ensures that the distance between $f(x)$ and $L$ will be less than our chosen epsilon. Does this set-up not imply continuity? Wikipedia says that c needs to be a "limit point" on our domain. Does that mean $f(c)$ has to exist or just that its limit has to exist?

Answer 1

In order for continuity at $c$ to hold, we require that $$\lim_{x \rightarrow c} f(x) = f(c)$$ While it may be the case that $\lim_{x\rightarrow c} f(x)$ exists, it is clearly not equal to $f(c)$, since $f(c)$ is itself undefined. This type of discontinuity is called a removable discontinuity, and in some ways is the "best" type of discontinuity to have, since it allows you to simply redefine a single point and end up with a continuous function.

EDIT using $\epsilon-\delta$ formulation: The definition of continuity at a point can also be written as

A function $f$ is continuous at a point $c$ if for all $\epsilon>0$, there exists a $\delta>0$ such that

$$|x-c| < \delta \implies |f(x) - f(c)| < \epsilon$$

But clearly we cannot apply this definition to the point $c$; if $f(c)$ is undefined, then so is $|f(x) - f(c)|$. You can choose any $\delta$ you like but the statement $|f(x) - f(c)|<\epsilon$ will always be false.

Answer 2

The definition of continuity is that $f(c)=L$. If $f(c)$ does not exist you cannot satisfy this. If $f(c)$ is undefined, you can define a new function $g$ by $$g(x)=\begin {cases} f(x)&x \neq c\\L&x=c \end {cases}$$. $g(x)$ will be continuous. We often call this a removable singularity because $g$ removes it. An example would be $f(x)=\frac {x-1}{x^2-1}$, which equals $x+1$ except at the point $x=1$ where it is not defined. Our $g(x)=x+1$ and is continuous at $1$.

Answer 3

To say that $f$ is continuous at $c$, $f(c)$ must exist. If $\lim_{x\to c} f(x)$ exists and equals $L$, then you can "extend" $f$ to make $f$ continuous at $c$ by defining $f(c)=L$.

I haven't looked at wiki article on continuity, but continuity doesn't require $c$ to be a limit point.

The notion of $\lim_{x\to c} f(x)$ only makes sense if $c$ is a limit point of the domain, but continuity doesn't necessarily require existence of this limit (Look carefully at definition of continuity). In fact, any function is continuous at isolated points in the domain.

Answer 4

If a function is continuous at every point in its domain, then it is a continuous function. The point $c$ in your example is not in the domain of your function and should not affect the continuity of your function.

Here is the definition in Wikipedia:

For any number $\varepsilon > 0$, however small, there exists some number $\delta > 0$ such that for all $x$ in the domain of $f$ with $x_0 − \delta < x < x_0 + \delta$, the value of $f(x)$ satisfies

$$f(x_{0})-\varepsilon <f(x)<f(x_{0})+\varepsilon.$$