AP Calculus AB and BC

Limit of arctan(x) at Negative Infinity

Q: Where does this limit appear outside graphing?

In improper integrals. \arctan x antidifferentiates \frac{1}{1+x^{2}}, so evaluating at a lower infinite endpoint means taking this limit. That gives \int_{-\infty}^{0} \frac{1}{1+x^{2}}\,dx = \frac{\pi}{2}, and \pi across the whole real line.

The limit of arctan x as x approaches negative infinity is negative pi over 2, roughly negative 1.5708. Arctangent is bounded below by that value and never reaches it, so y = negative pi over 2 is a horizontal asymptote on the left. At positive infinity the limit is pi over 2.

\lim_{x \to -\infty} \arctan x = -\frac{\pi}{2}

Settled by the range of the arctangent.

Where the floor comes from

Tangent repeats, so its inverse has to be built on a single branch. The branch chosen runs between the two vertical asymptotes nearest the origin, and on it tangent takes every real value exactly once.

\tan : \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \to (-\infty, \infty)

Inverting swaps domain and range, so $\arctan$ accepts any real number and returns an angle strictly inside that open interval. The interval answers the limit before any computation begins.

\arctan : (-\infty, \infty) \to \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)

Now read the limit as a question about angles. A hugely negative input asks which angle in that interval has a hugely negative tangent, and tangent runs down to $-\infty$ only as the angle drops toward $-\frac{\pi}{2}$ . The output is pressed against the bottom edge.

$x$	$\arctan x$
$-1$	$-0.7854$
$-10$	$-1.4711$
$-100$	$-1.5608$
$-1000$	$-1.5698$

The values fall steadily and stay above $-\frac{\pi}{2} \approx -1.5708$ , closing the gap without ever shutting it.

Why substitution fails, and why the limit has to exist

As at any infinite endpoint there is nothing to substitute, and unlike a rational function no indeterminate form shows up once you look. There is no $\frac{\infty}{\infty}$ and no $\infty - \infty$ , because a single bounded function has no competitor.

Existence is settled in advance. Arctangent decreases as $x$ decreases and is bounded below, and a monotone function with a bound in the direction it is heading must converge. All that remains is identifying the value, which the range supplies.

An unbounded input does not force an unbounded output

The instinct that a runaway input drags the output with it is what makes this limit feel wrong. $\arctan x$ , $\sin x$ , and $\frac{x}{\sqrt{x^{2}+1}}$ all accept every real input and return values confined to a fixed window.

Arctangent is odd, so the left end mirrors the right: $\arctan(-x) = -\arctan x$ converts the value $\frac{\pi}{2}$ at $\infty$ into $-\frac{\pi}{2}$ here. The graph carries two horizontal asymptotes, which is what Topic 1.15 questions probe.

The mistakes students make

The first is answering $-\infty$ , from picturing tangent instead of its inverse. Tangent does blow up near $-\frac{\pi}{2}$ , but that is the same relationship read the other way. Arctangent is the bounded direction.

The second is answering $-90$ . AP limits are in radians, so the exact answer is $-\frac{\pi}{2}$ , and a decimal is acceptable only alongside it.

The third is naming one horizontal asymptote when the question asks for all of them.

Limit	Value
$\lim_{x \to -\infty} \arctan x$	$-\frac{\pi}{2}$
$\lim_{x \to \infty} \arctan x$	$\frac{\pi}{2}$
$\lim_{x \to 0^{-}} \arctan\left(\frac{1}{x}\right)$	$-\frac{\pi}{2}$
$\lim_{x \to -\infty} \arctan\left(e^{x}\right)$	$0$

The last row is a reminder to read the inside first. There the inner function heads to 0 rather than to an infinite end, so the outer arctangent contributes nothing dramatic.

Not sure which technique a limit wants?

The Limit Method Chooser walks the decision from direct substitution through factoring, the conjugate, and L'Hopital, and says why each one applies or fails.

Frequently asked questions

Does arctan x ever equal negative pi over 2?

No. The range is the open interval, so the value is approached and never attained. If some input returned exactly $-\frac{\pi}{2}$ , then $\tan\left(-\frac{\pi}{2}\right)$ would have to equal that input, and tangent is undefined there.

How can one function have two horizontal asymptotes?

The two ends are independent questions, and each one records its own limit. So a function can have an asymptote on the left, on the right, on both, or on neither. Arctangent has $y = -\frac{\pi}{2}$ and $y = \frac{\pi}{2}$ , and $\frac{x}{\sqrt{x^{2}+1}}$ has $y = -1$ and $y = 1$ .

Where does this limit appear outside graphing?

In improper integrals. $\arctan x$ antidifferentiates $\frac{1}{1+x^{2}}$ , so evaluating at a lower infinite endpoint means taking this limit. That gives $\int_{-\infty}^{0} \frac{1}{1+x^{2}}\,dx = \frac{\pi}{2}$ , and $\pi$ across the whole real line.