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Assume we have a function $f$, say on $\mathbb{R}$, such that $f$ is continuously differentiable in all $x$ smaller than some given $x_0 \in \mathbb{R}$.

I am a bit confused about the connections of $\lim_{x \to x_0, x < x_0} f'(x)$ and the left derivative $f'_-(x_0)$ at $x_0$, since I could neither prove nor find a counterexample on these questions:

1) If $\lim_{x \to x_0, x < x_0} f'(x)$ exists, does so the left derivative at $x_0$ and are they equal?

2) If the left derivative exists, does the other limit exist and are they equal?

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  • $\begingroup$ At a minimum, you need continuity at $x_0$ for (1) to be true. $\endgroup$ – Thomas Andrews Jan 27 '16 at 17:48
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In the following I assume that $f$ is continuous.

The answer to 1) is yes, by the mean value theorem. For $h>0$ small enough $$ \frac{f(x_0-h)-f(x_0)}{-h}=f'(\xi_h),\quad x_0-h<\xi_h<x_0. $$ Letting $h\to0$ we get hat the left derivative exists at $x_0$ and equals $\lim_{x\to x_0^-}f'(x)$.

For 2) consider $f(x)=(x-x_0)^\alpha\sin\frac{1}{x-x_0}$ if $x\ne x_0$, $f(x_0)=0$ with $1<\alpha\le2$. Then $f$ is differentiable on $\mathbb{R}$, $C^\infty$ on $\mathbb{R}\setminus\{x_0\}$ and $\lim_{x\to x_0^\pm}f'(x)$ does not exist.

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