Wolfram|Alpha fails to recognize only 1-time differentiability at 0?

Questions
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1

Define f(x) = (x^2)sin(1/x) if x != 0 and define f(0) = 0.
If x != 0 then f’(x) = (2)(x)sin(1/x)-cos(1/x).
If x = 0 then f’(0) = lim ((x^2)sin(1/x)-0)/(x-0) as x->0 = lim (x)sin(1/x) as x->0 = 0.
So f(x) is at least 1-time differentiable at x=0.
Now lim f’(x) as x->0 = lim (2)(x)sin(1/x)-cos(1/x) as x->0 does not exist.
Hence, we cannot have lim f’(x) = f’(0) as x->0, i.e., f’(x) dose not continuous at x=0, which implies f’(x) is not differentiable at x=0, i.e., f(x) is only 1-time differentiable at x=0.
In WolframAlphan:
Inputting D[Piecewise[{{x^2 Sin[1/x], x != 0}, {0, x == 0}}], x] outputs Piecewise[{{-Cos[x^(-1)] + 2 x Sin[x^(-1)], x != 0}}, 0], which is correct math result.

But inputting D[Piecewise[{{-Cos[x^(-1)] + 2 x Sin[x^(-1)], x != 0}}, 0], x] outputs Piecewise[{{(-2 x Cos[x^(-1)] + (-1 + 2 x^2) Sin[x^(-1)])/x^2, x != 0}}, 0], which is incorrect math result; it does not show non-differentiability at x=0.

Also inputting D[D[Piecewise[{{x^2 Sin[1/x], x != 0}, {0, x == 0}}], x], x] again outputs the same incorrect math result.

This is the problem.

2

To prove this, simply do the following:

Sin[1/t] t^2 = Sin[1/t]/(1/t) t

Take the limit:

Limit[Sin[1/t]/(1/t) t, t -> 0] = Limit[Sin[1/t]/(1/t), t -> 0] Limit[t, t -> 0] = 1*0

And you’ll get your answer: 0

Also, interesting implications about the series of a bunch of reciprocals converging to a single point as they all approach the same pole.

3

I misread your statement, never mind the above post.