AP Calculus AB and BC

Limit of arctan(x) as x Approaches Infinity

Q: Does arctan x ever equal pi/2?

No. The range is the open interval, so \frac{\pi}{2} is approached and never attained. If some input gave exactly \frac{\pi}{2}, then \tan\left(\frac{\pi}{2}\right) would have to equal that input, and tangent is undefined there. The limit describes a ceiling, not a value on the graph.

Q: What is the limit as x approaches negative infinity?

It is -\frac{\pi}{2}. Arctangent is odd, meaning \arctan(-x) = -\arctan x, so the left end mirrors the right. The graph therefore has two distinct horizontal asymptotes, y = \frac{\pi}{2} and y = -\frac{\pi}{2}.

Q: Why does pi/2 keep appearing in improper integrals?

Because \arctan x is the antiderivative of \frac{1}{1+x^{2}}, and evaluating at the infinite endpoint means taking this limit. That is what makes \int_{0}^{\infty} \frac{1}{1+x^{2}}\,dx = \frac{\pi}{2}, and the full integral over the whole real line equal to \pi.

The limit of arctan x as x approaches infinity is pi over 2, about 1.5708. Arctangent is bounded: it rises toward the top of its range without ever reaching it, so y = pi over 2 is a horizontal asymptote on the right. At negative infinity the limit is negative pi over 2, giving a second asymptote.

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

Settled by the range of the arctangent.

Where the ceiling comes from

Tangent is not one-to-one, so the inverse is defined on a restricted branch. The branch chosen is the one running between the two vertical asymptotes closest to the origin, on which tangent climbs through every real value exactly once.

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

Inverting a function swaps its domain and its range, so arctangent accepts every real number and returns an angle strictly inside that open interval.

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

Now read the limit as a question about angles. Asking for $\arctan x$ with $x$ enormous asks which angle in that interval has an enormous tangent, and tangent blows up only as the angle approaches $\frac{\pi}{2}$ from below. So the output is forced up against that edge.

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

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

Why substitution fails, and why nothing is indeterminate

As at every infinite endpoint, there is no number to substitute. What makes this one different from a rational function is that no indeterminate form appears afterwards either. There is no $\frac{\infty}{\infty}$ and no $\infty - \infty$ , because no two quantities are competing.

In fact the existence of the limit is guaranteed before any value is computed. Arctangent is increasing everywhere and bounded above by $\frac{\pi}{2}$ , and an increasing function with a ceiling has to converge to some value at or below that ceiling. The remaining work is identifying which value, and the range settles that.

Unbounded input, bounded output

The hesitation students feel here is that an input running to infinity ought to drag the output with it. Nothing requires that. Arctangent, $\sin x$ , and $\frac{x}{\sqrt{x^{2}+1}}$ all take unbounded inputs and return values that stay inside a fixed window, here $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ , $[-1, 1]$ , and $(-1, 1)$ .

The mistakes students make

The first is answering $\infty$ , from confusing arctangent with tangent. Tangent does blow up as its input approaches $\frac{\pi}{2}$ , but that is the inverse function running the other way. Arctangent is the bounded one.

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

The third is forgetting the other end. Arctangent is odd, so the two ends give different horizontal asymptotes, and a graph question that asks for all of them wants both lines.

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

The last two rows are a favorite exam setup. Since $\frac{1}{x}$ runs to $\infty$ from the right of 0 and to $-\infty$ from the left, the composition $\arctan\left(\frac{1}{x}\right)$ has a jump discontinuity at $x = 0$ of height $\pi$ .

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 pi/2?

No. The range is the open interval, so $\frac{\pi}{2}$ is approached and never attained. If some input gave exactly $\frac{\pi}{2}$ , then $\tan\left(\frac{\pi}{2}\right)$ would have to equal that input, and tangent is undefined there. The limit describes a ceiling, not a value on the graph.

What is the limit as x approaches negative infinity?

It is $-\frac{\pi}{2}$ . Arctangent is odd, meaning $\arctan(-x) = -\arctan x$ , so the left end mirrors the right. The graph therefore has two distinct horizontal asymptotes, $y = \frac{\pi}{2}$ and $y = -\frac{\pi}{2}$ .

Why does pi/2 keep appearing in improper integrals?

Because $\arctan x$ is the antiderivative of $\frac{1}{1+x^{2}}$ , and evaluating at the infinite endpoint means taking this limit. That is what makes $\int_{0}^{\infty} \frac{1}{1+x^{2}}\,dx = \frac{\pi}{2}$ , and the full integral over the whole real line equal to $\pi$ .