AP Calculus AB and BC

Limit of arctan(1/x) at 0 from the Right

Q: Why is the arctangent bounded?

It inverts the tangent restricted to the interval from -\pi/2 to \pi/2. That restriction becomes the arctangent's range, so its outputs can never leave it.

The limit of arctan of one over x as x approaches zero from the right is pi over 2. The inner expression runs to positive infinity on that side, and the arctangent function levels off at pi over 2 as its input grows.

\lim_{x \to 0^+} \arctan\!\left(\frac{1}{x}\right) = \frac{\pi}{2}

Settled by tracking the inner expression, then the range of the arctangent.

Two steps, inside out

First, $1/x \to +\infty$ as $x \to 0^{+}$ . Then the arctangent of a large positive input approaches its horizontal asymptote.

\lim_{x \to 0^{+}} \arctan\!\left(\frac{1}{x}\right) = \lim_{u \to \infty} \arctan u = \frac{\pi}{2}

The value $\pi/2$ is never actually attained, since the arctangent's range is the OPEN interval from $-\pi/2$ to $\pi/2$ . A limit that is approached but never reached is still a perfectly good limit.

The mirror image on the left

As $x \to 0^{-}$ the inner expression runs to $-\infty$ and the limit is $-\pi/2$ . The two sides differ by $\pi$ , so the two-sided limit does not exist and the origin is a jump discontinuity for this function.

Notice this is a jump rather than an infinite discontinuity: both one-sided limits are finite, they just disagree.

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

Why is the arctangent bounded?

It inverts the tangent restricted to the interval from $-\pi/2$ to $\pi/2$ . That restriction becomes the arctangent's range, so its outputs can never leave it.

Does arctan(1/x) ever equal pi/2?

No. It gets arbitrarily close but never reaches it, because the arctangent's range is open at both ends.