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If you are given

$f(x):\Bbb{R} \rightarrow \Bbb{R}$ defined on $[c,\infty)$ and

$\displaystyle\lim_{x \to c^+}{f(x) = f(c)}$

Then, we say that $f(x)$ is "right continuous" at $c$

Is it the case that one of these must be true?

  • the slope approaches a finite value and thus $f(x)$ is "right differentiable"
  • the slope approaches an infinite value and thus $f(x)$ is not "right differentiable"

Is there any other (weird) possibility?

Henry Bigelow
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2 Answers2

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Consider $$f(x)=\begin{cases}x\sin\frac1x&x>0\\0&x=0\end{cases}$$ or for additional fun $$f(x)=\begin{cases}x\sin\frac1x&x\notin\Bbb Q\\0&x\in\Bbb Q\end{cases}$$

Hagen von Eitzen
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  • That is crazed! Is there some intuition for how this is possible? It seems to me that the delta epsilon argument for continuity would imply that, if the limit exists at all, it means the ratio of delta to epsilon converges to some value and the function is locally linear. – Henry Bigelow Oct 16 '18 at 23:11
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The cases you have considered correspond for example to

  • $f(x)=x^2$ for $x\ge 0$
  • $f(x)=\sqrt x$ for $x\ge 0$

As a different case let consider for $x\ge 0$

$$f(x)=x\cdot \sin \frac1x$$

with $f(0)=0$, then the right side derivative doesn't exist indeed we obtain

$$\lim_{x\to 0^+} \frac{x\cdot \sin \frac1x-0}{x-0}$$

refer also to the related

user
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  • Hi @gimusi, thanks for your answer. Can you possibly give an intuition for how this is possible? It seems to me that, if the limit exists, it implies that, if we consider the maximal value of delta that satisfies the given epsilon, and take the ratio of epsilon / delta, they must converge to some real value. And, why wouldn't this value be the slope of the function? – Henry Bigelow Oct 16 '18 at 23:13
  • @HenryBigelow No the limit $\lim_{x\to 0^+} \sin \frac1x$ doesn't exist. – user Oct 16 '18 at 23:14