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chech whether the function is differentiable at $x=0$

$$f(x)=\left\lbrace \begin{array}{cl} \arctan\frac{1}{\left | x \right |}, & x\neq 0 \\ \frac{\pi}{2}, & x=0\\ \end{array}\right.$$

I feel that this is differentable at given point but I am unable to proceed with it.

Steven Alexis Gregory
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    Note that $\arctan(1/|x|) = \frac{\pi}{2} - \arctan|x|$. Since $\arctan |x| \approx |x|$ near $x = 0$, this means that the graph of $f$ looks like a wedge near $x = 0$. – Sangchul Lee Aug 22 '15 at 04:22

2 Answers2

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As suggested in comments, if $x \not =0$, $\frac{f(x)-f(0)}{x-0}=\frac{-\arctan|x|}{x}$. Now if $x \to 0^{+}$, then $\lim\limits_{x \to 0^{+}}\frac{-\arctan|x|}{x}=\lim\limits_{x \to 0^{+}}\frac{-\arctan x}{x}=\lim\limits_{x \to 0^{+}}\frac{-\arctan x - (-\arctan(0)}{x-0}=(-\arctan)'(0)=-\frac{1}{0^{2}+1}=-1$, Now as $x \to 0^{-}$, $\lim\limits_{x \to 0^{-}}\frac{-\arctan|x|}{x}=\lim\limits_{x \to 0^{-}}\frac{-\arctan(-x)}{x}=1$ (Same way as before, and we use the chain rule, the derivative of $k(x)=\arctan(-x)$ is $\frac{-1}{x^{2}+1}$.) The limit is different at both sides, so $f$ is not differentiable at $0$

mich95
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Simply use the identity

$$\arctan (1/x)=\frac{\pi}{2}\text{sgn}(x)-\arctan(x)$$

Then, the limit of the difference quotient is

$$\lim_{h\to0}\frac{\arctan(1/|h|)-\frac{\pi}{2}}{h}=-\lim_{h\to0}\frac{\arctan(|h|)}{h} \tag 1$$

We see that the limit in $(1)$ does not exist since the limit from the right side does not equal the limit from the left side. In fact, a quick check using say L'Hospital's Rule shows that the limit from the right side is $-1$, while the limit from the left side is $1$.

We conclude, therefore, that $\arctan(1/|x|)$ is not differentiable at $x=0$.

Mark Viola
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