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Consider the function $f(x)$ defined by \begin{equation*} f(x) = \begin{cases} \tan(x/2 + \pi/4) - (\pi/4)/(1-2 x/\pi) & \text{if $x \ne \pi/2$,} \\ 0 & \text{if $x = \pi/2$}. \end{cases} \end{equation*}

Combining fractions and then using L'Hopital's rule (twice), it can be seen that this function is continuous at $\pi/2$.

Since the only discontinutites of this function are at multiples of $\pi/2$, can we conclude, without any further calculations, that the function is also differentiable at $\pi/2$?

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  • $\begingroup$ Yes. $\,\,\!\!$ $\endgroup$ – Berci Mar 24 '15 at 0:33
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    $\begingroup$ I don't think so. Differentiability of a function at a point ensures its continuity at that point but the converse of this theorem is not always true. $\endgroup$ – Prasun Biswas Mar 24 '15 at 0:35
  • $\begingroup$ What is $$\lim_{h \to 0} \frac{f(\frac{\pi}{2}+h)}{h}$$? $\endgroup$ – Oria Gruber Mar 24 '15 at 0:39
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make a change of variable $$\begin{align}x = \pi/2 + 2h, f(\pi/2 + 2h) &= \tan (\pi/2 + h) - (\pi/4)\left(1 - \frac{2}{\pi}(\pi/2 + 2h) \right)\\ &= -\tan h + h = -\frac 13 h^3+\cdots\end{align}$$ therefore $f$ is differentiable at $x = \pi/2$ and its derivative is zero.

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