ask question about left derivative and limit of derivative from left-hand

Question

I know the definition of left derivative. If function f is differentiable and the limit of its derivative from left-hand of point x exists then it equals to left derivative of point x . This can be proven by using mean value theorem . my question is when the limit of derivative from left-hand of x doesn't exist why can left derivative exist . I know the example is $f(x)=x^2 \sin(\frac{1}{x})$.but I want to know why the proof by using mean value theorem fails

by mean value theorem there is c between [x,a],s.t. $\frac{f(x)-f(a)}{x-a}=f'(c)$, take limit on both side as x goes to a- , then $\lim \limits_{x \rightarrow a-} \frac{f(x)-f(a)}{x-a}=\lim \limits_{x \rightarrow a-}f'(c)$ . if $\lim \limits_{x \rightarrow a-}f'(c)=A$ , then I can conclude left derivative at a=$\lim \limits_{x \rightarrow a-} \frac{f(x)-f(a)}{x-a}=A$

but if $\lim \limits_{x \rightarrow a-}f'(c)$ doesn't exist ,there still exists c so that the equality $\frac{f(x)-f(a)}{x-a}=f'(c)$ and $\lim \limits_{x \rightarrow a-} \frac{f(x)-f(a)}{x-a}=\lim \limits_{x \rightarrow a-}f'(c)$ still hold,then $\lim \limits_{x \rightarrow a-}f'(c)$ doesn't exist can conclude left derivative=$\lim \limits_{x \rightarrow a-} \frac{f(x)-f(a)}{x-a}$ doesn't exist .

where am I wrong

Answer 1

by mean value theorem there is c between [x,a],s.t. $\frac{f(x)-f(a)}{x-a}=f'(c)$, take limit on both side as x goes to a- , then $\lim \limits_{x \rightarrow a-} \frac{f(x)-f(a)}{x-a}=\lim \limits_{x \rightarrow a-}f'(c)$ . if $\lim \limits_{x \rightarrow a-}f'(c)=A$ , then I can conclude left derivative at a=$\lim \limits_{x \rightarrow a-} \frac{f(x)-f(a)}{x-a}=A$

If this is a proof about a general function $f$ then the function may not satisfy the conditions for the MVT, which require the function to be continuous on $[x,a]$ and for $f'(\xi)$ to exist for every $\xi$ in $(x,a).$

But consider the specific example you raised,

$$ f(x) = \begin{cases} x^2 \sin(1/x) & x \neq 0, \\ 0 & x = 0. \end{cases} $$

Although this function does satisfy the conditions of the MVT, the proof fails for $a = 0$ because it does not consider how a limit is actually defined. What you have shown is that for any $\epsilon,$ the interval $(x,a)$ is a left-side neighborhood of $a$ within which there exists one value of $\xi,$ namely $\xi = c$ found by the MVT, for which $\lvert f'(\xi) - A \rvert < \epsilon.$ But in order to show a limit you need to demonstrate the existence of a left-side neighborhood of $a$ in which every value of $\xi$ satisfies $\lvert f'(\xi) - A \rvert < \epsilon.$ This requirement is spoiled for this function because in any interval $(x,0)$ there are values of $\xi$ for which $f'(\xi) > \frac12$ and other values of $\xi$ for which $f'(\xi) < -\frac12.$