# Question on one-sided derivatives

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?

• At a minimum, you need continuity at $x_0$ for (1) to be true. – Thomas Andrews Jan 27 '16 at 17:48

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 $$\frac{f(x_0-h)-f(x_0)}{-h}=f'(\xi_h),\quad x_0-h<\xi_h Letting $$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.