I was reading this question.
The simplest examples of continuous functions, with discontinuous derivatives in some point, are usually of the form:
f(x)={x2sin(1/x)if x≠00if x=0.
The derivative of f is
f′(x)={2xsin(1x)−cos(1x)if x≠00if x=0,
The derivative is discontinuous because the limit of cos(1x) does not exist for x→0.
Is there an example where the derivative is still discontinuous but with existing limit?
Thanks
Answer
Suppose f is differentiable in some neighborhood (x−δ,x+δ) of x, and lim exists. Define y:(x-\delta,x+\delta)\rightarrow\mathbb{R} such that y(t) is strictly between x and t and
f(t)-f(x) = f'(y(t))(t-x)
for every t\in(x-\delta,x+\delta). The existence of such a function is guaranteed by the Mean Value Theorem. Since y(t) is between x and t for every t, this implies that \lim\limits_{t\rightarrow x}{y(t)} = x, and since y(t)\ne x for t\ne x as well, we have \lim\limits_{t\rightarrow x}{f'(y(t))}=\lim\limits_{s\rightarrow x}{f'(s)} by the composition law (think of s = y(t) in this substitution). This implies that
f'(x) = \lim\limits_{t\rightarrow x}{\frac{f(t)-f(x)}{t-x}} = \lim\limits_{t\rightarrow x}{f'(y(t))} = \lim\limits_{t\rightarrow x}{f'(t)},
i.e. f' is continuous at x.
Remark: Typically, the composition law is phrased as follows: if \lim\limits_{x\rightarrow c}{g(x)} = a and f is continuous at a, then \lim\limits_{x\rightarrow c}{f(g(x))} = \lim\limits_{u\rightarrow a}{f(u)}. In our problem, we obviously cannot assume f' is continuous at x, since that is what we are trying to show. However, the above conclusion still holds if we merely require that g(x)\ne a if x\ne c in some neighborhood of c. A proof of this can be found here (look for "Hypothesis 2").
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