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

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

• MVT theorem is only for one way implication. How would you use MVT to prove that if left derivative exists then the limit of the left derivative exists? In fact left derivative may not even exist in any interval around the point. Jan 6 at 9:02

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 $$(x,a).$$
$$f(x) = \begin{cases} x^2 \sin(1/x) & x \neq 0, \\ 0 & x = 0. \end{cases}$$
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 $$f'(\xi) < -\frac12.$$