In the very beginning, I'm going to refer to similar posts dealing with proofs of continuity(discontinuity) of a function at a point or on an open interval:

Proving that a specific function isn't continuous

Help with epsilon-delta proof that 1/(x^2) is continuous at a point.

My problem:

Prove that $f:\mathbb R\to\mathbb R$ $$f:=\begin{cases}3^x,\;\;\;\;\;x<0\\3^{-x+1},x\geq0\end{cases}$$ is discontinuous at $x=0$ by the Cauchy definition.

By the definition (source: Continuity &Limits/Neprekidnost i limesi):

Let $I\subseteq\mathbb R$ be an open interval. Function $f$ is continuous at a point $c\in I$ iff: $$(\forall\varepsilon>0)(\exists\delta>0) s.t. x\in I,|x-c|<\delta\implies|f(x)-f(c)|<\varepsilon$$ Since I have to prove discontinuity at $x=0$, I negated the statement and plugged $x=0$ into it: $$(\exists\varepsilon>0)(\forall\delta>0)s.t.x\in I ,|x|<\delta\;\land\;|f(x)-f(0)|\geq\varepsilon$$

$$\iff(\exists\varepsilon>0)(\forall\delta>0)s.t.x\in I,|x|<\delta\;\land\;|f(x)-3|\geq\varepsilon$$



I'm a bit confused because I have only seen such proofs of Dirichlet's function and certain step-functions. First thing I did was finding $$\lim_{x\to 0^+}f(x)=1\;\&\;\lim_{x\to 0^-}f(x)=3$$ and we cannot extend the function by some continuous function from that point. Then I decided to solve the inequality to find sufficiently small $\varepsilon$, but I didn't get far from what I already had:$$|f(x)-3|\geq\varepsilon$$ What should I do next if the function is defined by $2$ formulae on $2$ disjoint intervals?


There is no limit $L$ that works. Suppose we take some possible limit $L < 2.$ No matter how small $\delta > 0$ is, there are points with $-\delta < x < 0$ such that $f(x) > \frac{5}{2},$ so violates the limit condition for any $0 < \varepsilon < \frac{1}{2}$

Similar for $L \geq 2,$ switching to $0 < x < \delta$

  • $\begingroup$ Can I conclude that at the end of the proof by Cauchy definition? Is it concise enough and legitimate in the context of this peculiar definition? $\endgroup$ – ms._VerkhovtsevaKatya Jan 26 at 16:44
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    $\begingroup$ @VerkhovtsevaKatya yes, this is correct. I think you are getting trouble from trying to expand everything into logic symbols, also you don't have a good visual sense. I recommend printing pdf's of graph paper (or buy graph paper) and making a habit of plotting the simpler functions you come across; telling a computer to plot something and staring at the result is not the same incompetech.com/graphpaper $\endgroup$ – Will Jagy Jan 26 at 17:14
  • $\begingroup$ thank you! I appreciate your advice. $\endgroup$ – ms._VerkhovtsevaKatya Jan 26 at 17:19
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    $\begingroup$ @VerkhovtsevaKatya back again. People my age graphed functions by hand. Meanwhile, note that "belletristic" is not a noun, it is an adjective. The original phrase: en.wikipedia.org/wiki/Belles-lettres $\endgroup$ – Will Jagy Jan 26 at 22:07

This is a long comment.

Just to clarify the negation of logical statement is wrong. You should emphasize that there exists some $x$ with $|x|<\delta$ such that $|f(x) - f(0)|\geq \epsilon$. Your negation seems to imply that this holds for all $x$ with $|x|<\delta$. Unless you are an expert in logic, it is better to explain / write definitions involving $\epsilon, \delta$ in your language of communication. Too much symbolism is a sure way to mess up your understanding.

  • $\begingroup$ thank you for commenting. Some things written by our assistant are suspicious, but therefore I ask. Thank you once again. I have an exam today so every feedback is helpful to me. $\endgroup$ – ms._VerkhovtsevaKatya Jan 27 at 7:29

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