Show $f(x)=\frac{1}{|x|}$ is discontinuous at x=0

Question

I am struggling to explicitly show that the function is discontinuous at $x=0$ since the usual technique of finding two sequences $x_n$,$y_n$ where $x_n\rightarrow0$, $y_n\rightarrow0$ but $f(x_n)$ and $f(y_n)$ tend to different values, doesnt seem to work as all seuqnces tend to $+\infty$.

My only other thought would be that since $f(0)$ doesn't exist, then from this $f(x)$ cannot be continuous. But I'm unsure if this would suffice as an answer.

The function $\frac{1}{|x|}$ is not defined at $x=0$, and so you cannot ask whether it is continuous there. It is continuous on $\mathbb{R}\setminus {0}$. — Damien, May 21 '18 at 11:33
Possible duplicate of Can we talk about the continuity/discontinuity of a function at a point which is not in its domain? — Michael Hoppe, May 21 '18 at 12:36
Roughly speaking, the following ensemble of symbols and words:
‘$1/|x| \colon\mathbb R^*\to\mathbb R$ is not continuous in $0$.’

is no statement, i.e., neither true nor false. — Michael Hoppe, May 21 '18 at 12:40

score 2 · Answer 1 · answered May 21 '18 at 11:21

If my students dare to say that this function is discontinuous at $0$, they will miserably fail. But we all know that the definition of discontinuity is somehow vague.

Many mathematicians clearly state that a function can be discontinuous only at points of the domain. For these people, your work is useless, since your function is continuous throughout the domain of definition.
Other mathematicians say that a function is always discontinuous at points that do not belong to the domain of definition, but they are accumulation points for the domain. Indeed, since $f(x_0)=\lim_{x \to x_0}f(x)$ is the characterizing property of continuous functions, it is violated as soon as $f(x_0)$ does not exist. Even in this case your work is useless, since your function is automatically continuous.

I won't write here a long discussion about the "best" definition of discontinuity, since it has been done several times in the past.

score 0 · Answer 2 · answered May 21 '18 at 11:19

0

Let $\displaystyle x_n=\frac{1}{n}$ then $\displaystyle x_n \underset{n \rightarrow +\infty}{\rightarrow}0$ and $$ f\left(x_n\right) \underset{n \rightarrow +\infty}{\rightarrow}+\infty $$

Hence $f$ is not continuous at $x=0$

answered May 21 '18 at 11:19

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But $f(x)=\frac{1}{|x|}$ is not definied for $x=0$ so your question is senseless – Dr. Sonnhard Graubner May 21 '18 at 11:22

Show $f(x)=\frac{1}{|x|}$ is discontinuous at x=0

2 Answers2