# Discontinuity of the function at a point by the Cauchy definition

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

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

and

$$x\in\langle-\delta,\delta\rangle$$

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 $$0 < x < \delta$$

• 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? – ms._VerkhovtsevaKatya Jan 26 at 16:44
• @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 – Will Jagy Jan 26 at 17:14
• thank you! I appreciate your advice. – ms._VerkhovtsevaKatya Jan 26 at 17:19
• @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 – 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.

• 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. – ms._VerkhovtsevaKatya Jan 27 at 7:29