y = \arctan\dfrac{1}{x}

Graph of the Function $y = \arctan\dfrac{1}{x}$

$y = \arctan\dfrac{1}{x}$ is the inverse tangent applied to a reciprocal. Its values stay between $-\dfrac{\pi}{2}$ and $\dfrac{\pi}{2}$ , and yet it jumps by $\pi$ at the origin, which makes for unusually instructive behavior.

Domain and range

The domain is $x \neq 0$ . Since $\dfrac{1}{x}$ is never $0$ , neither is $\arctan\dfrac{1}{x}$ , so the branch with $x > 0$ lies in $\left( 0, \dfrac{\pi}{2} \right)$ and the branch with $x < 0$ in $\left( -\dfrac{\pi}{2}, 0 \right)$ .

The jump at the origin

Approach	$\dfrac{1}{x}$	$y$
$x \to 0^{+}$	$\to +\infty$	$\to \dfrac{\pi}{2}$
$x \to 0^{-}$	$\to -\infty$	$\to -\dfrac{\pi}{2}$

Both one-sided limits exist but differ, so this is a jump discontinuity, of size exactly $\pi$ . No value assigned at $x = 0$ could make the function continuous.

Because the values stay bounded, the $y$ -axis is not a vertical asymptote. The difference between a discontinuity where the function blows up and one where it merely jumps is plain to see here¹.

Symmetry and asymptote

From $f(-x) = -f(x)$ the function is odd, with rotational symmetry about the origin. As $x \to \pm\infty$ we have $\dfrac{1}{x} \to 0$ and so $y \to 0$ , making the $x$ -axis a horizontal asymptote, approached from above on the right and from below on the left.

Monotonicity

The chain rule gives a derivative that simplifies neatly.

\begin{align*} y' &= \frac{1}{1 + \left(\dfrac{1}{x}\right)^2} \cdot \left(-\frac{1}{x^2}\right) \\ &= -\frac{1}{1 + x^2} \end{align*}

The simplified right-hand side accepts $x = 0$ and takes the value $-1$ there. The function itself jumps, yet the formula for its derivative appears to run smoothly across the origin. Both branches approach the origin with slope $-1$ and are simply offset from each other by $\pi$ .

Relation to the inverse tangent

Because this derivative is the exact negative of $\dfrac{1}{1+x^2}$ , the derivative of $\arctan x$ , the sum $\arctan x + \arctan\dfrac{1}{x}$ is constant on each branch. Evaluating at one point on each gives the following.

\arctan x + \arctan\frac{1}{x} = \begin{cases} \dfrac{\pi}{2} & (x > 0) \\ -\dfrac{\pi}{2} & (x < 0) \end{cases}

This graph is therefore the continuous curve $\arctan x$ turned upside down and shifted up by $\dfrac{\pi}{2}$ on the right and down by $\dfrac{\pi}{2}$ on the left. The two shifts differ by $\pi$ , and that difference is precisely the jump at the origin.

Concavity

The second derivative is $y'' = \dfrac{2x}{(1+x^2)^2}$ , positive for $x > 0$ and negative for $x < 0$ , so the right branch is concave up and the left branch concave down. The sign changes at $x = 0$ , which lies outside the domain, so there is no inflection point.

Two conventions for the inverse cotangent

There are two competing definitions of $\operatorname{arccot}$ .

Convention	Formula	Range	Continuity
Principal value in $(0, \pi)$	$\dfrac{\pi}{2} - \arctan x$	$(0, \pi)$	continuous on the whole line
Through the reciprocal	$\arctan\dfrac{1}{x}$	$\left( -\dfrac{\pi}{2}, \dfrac{\pi}{2} \right)$ without $0$	jumps at the origin

The second is this graph. It differs from the first by $\pi$ for $x < 0$ , so it is always worth checking which convention a textbook or a computer algebra system has adopted.

Applications

When a program recovers an angle from coordinates, using $\arctan\dfrac{y}{x}$ alone cannot distinguish the first quadrant from the third, or the second from the fourth. That is why most languages provide a separate two-argument function, atan2, which avoids this discontinuity and returns the full range $(-\pi, \pi]$ ². This graph makes the need for it visible at a glance.

Classification of discontinuities, Wikipedia
atan2, Wikipedia

Graph of the Function y=arctan⁡1xy = \arctan\dfrac{1}{x}y=arctanx1​