Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How to show $\lim \limits_{x \rightarrow c_{-}} f(x) \neq \lim \limits_{x \rightarrow c_{+}} f(x)$ imply $f: \mathbb R \rightarrow \mathbb R$ is discontinuous at $c$ ?

I know that $f$ cannot have two different limits at $c$, since this is causing a contradiction.

Also if I draw $f$ under the given assumption I see there is a jump on the graph, which intuitively imply discontinuity.

How can I prove this is numbers ?

share|cite|improve this question
may be you should see what does it mean to say $\lim_{x \rightarrow c_{+}} f(x)$ and $\lim_{x \rightarrow c_-} f(x)$ – Praphulla Koushik Jun 11 '14 at 10:07
It is easier to prove that$\lim\limits_{x\rightarrow c_-}{f(x)}=\lim\limits_{x\rightarrow c_+}{f(x)}$ given the function is continuous at $c$. – fermesomme Jun 11 '14 at 10:40
up vote 3 down vote accepted


$$\lim \limits_{x \rightarrow c_{-}} f(x)= M$$

$$ \lim \limits_{x \rightarrow c_{+}} f(x) = L$$

Without loss of generality let $M > L$ (otherwise repeat the same arguments, etc.)

These statements mean that there exists a $\delta_1, \delta_2$ so that for any $\epsilon$ the following holds:

$$c < x < c + \delta_1 \implies |f(x) - M| \lt \epsilon$$

$$c - \delta_2 \ < x < c \implies |f(x) - L| \lt \epsilon$$

Take $\delta = \min(\delta_1, \delta_2)$ and $\epsilon = \frac{M-L}{2}$.

Combining the above statements, for $0 < |x-c| < \delta$, it must be that both $|f(x) - M| < \frac{M-L}{2}$ and that $|f(x) - L| < \frac{M-L}{2}$.

Can you arrive at a contradiction from this? It is essentially the same from here as the proof that a function cannot have two different limits at a point.

share|cite|improve this answer
Thank you, I use the trinagle inequality to get a contradiction with $|M-L| = |M-f(x)+f(x)-L| \le |f(x) - M| + |f(x) - L|$ – user111854 Jun 11 '14 at 11:11
  1. Write the definition of what $$\lim_{x\to c_-} f(x) = a$$ means.
  2. Do the same with the other limit.
  3. write the definition of the continuity of $f$ at $c$.
  4. Compare what you have and comment on it.

I cannot promise you that you will get the answer right away, but until you do these things, it is very hard for us to help you.

share|cite|improve this answer

Perhaps this perspective will help you.

The statement you seek to prove is equivalent to (its contrapositive):

  • If $f$ is continuous at $c$, then $ \lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x).$

But that comes from the definition of continuity of $f$ at $c$.

share|cite|improve this answer
Since the statement you wrote is exactly equivalent to the statement OP wants to prove, I highly doubt that he is allowed to assume this statement is true... Still, this would be the easiest way of proving what he needs...:D – 5xum Jun 11 '14 at 10:19
Yes 5xum - I agree with you :-) – Conan Wong Jun 11 '14 at 10:20

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.