@Thomas But I think the hypothesis did not say that...

It does. Read it again.

Not in the first sentence. After 'Show that'.

@Thomas Umm, the hypothesis says $\lim_{x\to a}f'(x)$ exists but not $\lim_{x\to b}f'(x)$ exists... And in my example $b=0$ while $a<0$.

For this you shuold consult the first sentence.($C^1((a,b])$ with an interval closed on the rhs.

@JoséCarlosSantos thanks, but no.

@Thomas I'm confused... That $f$ is continuous on $[a,b]$ and differentiable on $(a,b]$ does not mean $f'$ is continuous at $b$.

The confusion comes from the fact that you created a "counterexample" on the uncritical side of the internval, which escaped me. $f$ is assumed to be differentiable in $b$, but not in $a$. There is nothing to show in $b$, and your example does not work cause the function is not differentiable in $b$ (and my first statement is still true, even though I've looked at the wrong end of the interval. If $f$ is differentiable in $b$ the corresponding limit exists).

@Thomas And yes the confusion comes from the fact that I think that, in some sense, continuity of $f'$ at $a$ cannot lead to the continuity of $f'$ on the whole interval. For my example, $f$ is differentiable at $b=0$ and $f'(0)=0$, as you can check by the definition.

Calculate $f^\prime$. You'll get a $-\sin(\frac{1}{x})$ term, which is not continuos in $x=0$. At both ends of the interval the assumptions state that this limit exists. Differentiability alone is not sufficient.

@Thomas No. For $f'(0)$ you cannot use the usual formulas because $x^2\sin(1/x)$ is undefined at $x=0$. (And $f$ is a continuous extension of $x^2\sin(1/x)$.)

By definition $f'(0)=\lim_{x\to 0}(f(x)-f(0))/x=\lim_{x\to0} x\sin(1/x)=0$.

$f$ is assumed to be differentiable on $(a,b]$. The exercise says that, if the limit of $f^\prime$ at the open side of the interval exists, then $f^\prime$ is continuous there. Once again, your example does not fulfil this assumption.

Why? In fact $f'$ is an elementary function on $[a,0)$.

Write down $f^\prime$. You get $2x\sin(x) - \sin(1/x)$ The limit of this function at $x=0$ does not exist.

the first term converges to $0$ and the second term oscilates between $-$ and $1$, so no limit.

But it need not exist. Only the limit at $x=a$ need to exist.

Please. The exercise is about the behavior at the open end of the interval. If $a<0$ then for your example there is nothing to show.

In my example, $a<b=0$. $f'$ behaves well around $a$ but not $b$, which is exactly what I want, since the exercise actually states that, roughly, if the derivative is continuous at $a$ then it would be continuous at $b$ (and indeed any other point in the interval) as well. What do you mean by nothing to show?

Ok. last try. The question is about the behaviour in $a$ and the relevant claim is about the behaviour in $a$, which is the open end of the interval. Sorry, but I really don't see how to explain this to you if you are constantly ignoring this.

I think I can now see your point.

I'm not ignoring this, but the conclusion is simply false, it states that $f'$ is continuous on the whole interval!

The authors made the mistake of writing $C^1([a,b])$

But the behavior in $b$ is not what they are actually intending to look at.

So it's wrongly put. But the obviously intended behaviour is to conclude something about the behvior at $a$, from an assumption made about the behavior at $a$

Sorry for the misunderstanding.

I'll delete my comments.

Yes this is what I'm asking... Basically I'm self-studying, so I can't know whether I'm correct.

Well, now the doubt is cleared and thanks for paying attention anyways

Fine. Good luck.