I had to test whether or not the function $$x^2\cos\frac{\pi}{2x}$$is derivable at $x=0$, so when I checked for $x>0+$ it turned out a bit disaster , can anyone just help me to figure out my mistake cause math cannot be wrong....(sorry if the question is trivial for most of the experts here)(thanks in advance)
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Andrei
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Your question is not clear. Try explaining better what you are not understanding and write the math in the question using MathJax – Davi Barreira Jul 03 '20 at 17:49
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Very important point: If a function is defined piecewise, i.e., there is a separate rule for $f(0)$, then you must use the limit definition.
You started to do that but never finished. What is $$\lim_{h\to 0} h\cos\big(\frac{\pi}{2h}\big)?$$ Hint: Note that $\left|h\cos(\frac{\pi}{2h})\right|\le |h|$.
Your other approach amounts to asserting that $f'(0) = \lim\limits_{x\to 0} f'(x)$, and this will only be valid if in fact the derivative is continuous. Indeed, when you do this problem correctly, you see that $f'(0)=0$ and $\lim\limits_{x\to 0} f'(x)$ does not exist. There are lots of functions that are differentiable with discontinuous derivative. (Students often get confused because of the theorem that a differentiable function is always continuous. But this says nothing about the derivative's being continuous!)
Ted Shifrin
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The issue is with the calculation of the derivative of $1/x$. The formula that you use is valid everywhere, except at $0$. So your calculations are correct everywhere, but not at $0$.
Andrei
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