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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 ?

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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 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 $c$. –  Hat Man Jun 11 at 10:40

3 Answers 3

up vote 3 down vote accepted

Let:

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

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

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.

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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 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.

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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 $c$.

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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 at 10:19
    
Yes 5xum - I agree with you :-) –  Conan Wong Jun 11 at 10:20

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