Domain of arctan(1/x)
I had this as part of a question in an exam. And, I reasoned, even when it's arctan(1/0) (undefined), it is pi/2. And, so I said, domain belongs to all Real Numbers. Why isn't it this
$\endgroup$ 52 Answers
$\begingroup$$$\lim_{x\to0^+}\dfrac1x=+\infty\neq\lim_{x\to0^-}\dfrac1x=-\infty,\qquad\lim_{y\to+\infty}\arctan(y)=\dfrac\pi2\neq\lim_{y\to-\infty}\arctan(y)=-\dfrac\pi2$$ Since the left limit differs from the right limit, the limit does not exist. To exist, the two limits must be equal and finite.
$\endgroup$ $\begingroup$The function $f(x) = \arctan(1/x)$ is undefined at $x=0$, you cannot assume it is $\pi/2$ at such value of $x$. The domain of $f$ is $\mathbb{R} \setminus \{0\}$.
$\endgroup$
