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We have a function defined as $f:[a,b]\to\Bbb R$ such that

$$f(x)=\begin{cases}2,&\text{if } x\in(a,b]\\1,&\text{if }x=a\end{cases}$$ with $a<b$.

Notice that because the function is defined in a closed interval the limit

$$f'(a)=\lim_{x\underset{>}{\to} a}\frac{f(x)-f(a)}{x-a}=+\infty$$

exists because it is defined only from the right side. So, why can't we conclude that differentiability does not imply continuity?

P.S.: what a dumb... infinity is not a point of $\Bbb R$ so $f'(a)=\infty$ is not on the range of the derivative.

Karlo
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Masacroso
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    The function is neither continuous nor differentiable at $x = a$, and it is both continuous and differentiable at all other points. What exactly are you getting to, here? – Arthur Jun 10 '16 at 16:26
  • Why is not differentiable at $a$? I showed in the body of the question that the limit that defined the differentiability at the point exist. – Masacroso Jun 10 '16 at 16:27
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    Is the function differentiable at $1$ in the first place? No, it isn't because the limit is not finite. – egreg Jun 10 '16 at 16:28
  • And what is the reason then @egreg? This is exactly what Im asking. – Masacroso Jun 10 '16 at 16:29
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    It should also be noted that the notion of end-point differentiability is a subtle one. $x \mapsto |x|$ is not differentiable at $0$, but if the domain is restricted to $[0,1]$, say, it is differentiable from the right. This is why my preference is to define the derivative so that $f$ being defined in a neighbourhood of $c$ is necessary. – MathematicsStudent1122 Jun 10 '16 at 16:30
  • @MathematicsStudent1122 The concept indeed only makes sense for functions defined over an interval so there's “nothing” outside it. – egreg Jun 10 '16 at 16:40
  • For any interval $[a,b]\subset\mathbb R$, if $f$ is a differentiable function $[a,b]\to\mathbb R$ then also $f'$ is a function $[a,b]\to\mathbb R$. This is not the case with your example. – MPW Jun 10 '16 at 16:52
  • There is no requirement that the domain of a differentiable function be an open set. There are perfectly well-defined differentiable functions whose domain is $\mathbb Q$ for example. The derivative exists if the limit exists, and "$x\to a$" is understood to mean that $x$ approaches $a$ through points in the domain of the function. – MPW Jun 10 '16 at 16:55
  • @MPW That is true, but there are more advanced issues also. See: http://math.stackexchange.com/questions/126176/differentiablility-over-open-intervals – MathematicsStudent1122 Jun 10 '16 at 19:08

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The problem of being at a boundary is not relevant. The function $$ f(x)=\begin{cases} -1 & \text{if $x<0$} \\[4px] 0 & \text{if $x=0$} \\[4px] 1 & \text{if $x>0$} \end{cases} $$ satisfies $$ \lim_{x\to0}\frac{f(x)-f(0)}{x-0}=\infty $$ but this doesn't make it differentiable at $0$. All definitions I know ask for the limit $$ \lim_{x\to a}\frac{f(x)-f(a)}{x-a} $$ to exist and be finite.

In certain cases, such as $f(x)=\sqrt[3]{x}$, the limit is infinite, but the tangent exists as well. In other cases, like the one presented above, the tangent surely doesn't exist.

So there's no point in making a definition for the derivative that allows for “infinite values”, particularly in view of the concept that the derivative allows for approximating the function with a linear function in the vicinity of the point, which is not possible when the tangent is vertical.

See also Why not define infinite derivatives?

Community
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egreg
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  • Oh, ok... Thank you very much. Yes, this make sense through the different definitions of different books that I know. But I'll need to ask to be sure. In the book Im reading I think there is no explicit reference to the finiteness but in some obscure way it make sense cause in the definition of limit the infinite limit are defined as a special case because they represent divergence. – Masacroso Jun 10 '16 at 16:45
  • @Masacroso Possibly it is not enough emphasized, but it is quite certainly assumed that the limit must be finite. – egreg Jun 10 '16 at 16:47
  • Well, infinite is not a point of $\Bbb R$, after all the derivative can not be defined as a function if is infinite, Im right? – Masacroso Jun 10 '16 at 17:06
  • @Masacroso That's one way to see the business. – egreg Jun 10 '16 at 17:08
  • In fact, there is quite a bit work in real analysis that has been done with infinite derivatives, even as far back as the late 1800s, as the google searches I suggested in my comment at Why not Define Infinite Derivatives? will show. Of course, this is not something that belongs in a calculus class (but plenty of real analysis texts touch on this topic). However, it would help to distinguish between conventions used in lower level mathematics classes and what actually exists in the literature. – Dave L. Renfro Jun 10 '16 at 17:50