AP Calculus AB and BC

Limit of arctan x / x as x Approaches 0 Is 1

Q: What is the limit of arctan x / x as x approaches 0?

It is 1, since the derivative of \arctan x at 0 is \frac{1}{1+0} = 1.

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

It is 0. Arctangent is bounded by \frac{\pi}{2} while the denominator grows without bound.

Q: Which other functions give 1 when divided by x near 0?

Any function with value 0 and slope 1 at the origin: \sin x, \tan x, \arcsin x, \arctan x, e^{x}-1 and \ln(1+x).

The limit of arctan x over x as x approaches 0 is 1. The form is 0 over 0, and one pass of L'Hopital's rule gives 1 over 1 plus x squared, which is 1 at the origin. The underlying reason is that arctan x behaves like x for small x.

\lim_{x \to 0} \frac{\arctan x}{x} = 1

Settled by L'Hopital's rule, or the Maclaurin series.

One pass of L'Hopital

Substitution gives $\frac{0}{0}$ , which the rule accepts, and the derivative of arctangent is algebraic.

\lim_{x \to 0}\frac{\arctan x}{x} \;\overset{\text{H}}{=}\; \lim_{x \to 0}\frac{\frac{1}{1+x^{2}}}{1} = \frac{1}{1+0} = 1

The series says the same thing

The Maclaurin series for arctan x starts x minus x cubed over 3, so dividing by x leaves 1 minus x squared over 3, which is 1 at the origin. The series also shows the approach is from below for small x.

Why arctan behaves like x near zero

A differentiable function is well approximated near a point by its tangent line there. At the origin arctangent passes through $0$ with slope $1$ , so its tangent line is $y = x$ , and dividing by $x$ compares the function with that line.

This is why $\sin x$ , $\tan x$ , $\arcsin x$ , $\arctan x$ , $e^{x} - 1$ and $\ln(1+x)$ all give $1$ when divided by $x$ near the origin: every one of them has value $0$ and slope $1$ there.

The mistakes students make

Confusing $\arctan x$ with $\cot x$ or with $\frac{1}{\tan x}$ . Those are the same as each other and both different from the inverse function.
Using the limit at infinity instead. $\frac{\arctan x}{x} \to 0$ as $x \to \infty$ , because arctangent flattens out at $\frac{\pi}{2}$ while $x$ keeps growing.
Working in degrees. As with every trigonometric limit, the derivative formula assumes radians.

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

What is the limit of arctan x / x as x approaches 0?

It is $1$ , since the derivative of $\arctan x$ at $0$ is $\frac{1}{1+0} = 1$ .

What is the limit as x approaches infinity instead?

It is $0$ . Arctangent is bounded by $\frac{\pi}{2}$ while the denominator grows without bound.

Which other functions give 1 when divided by x near 0?

Any function with value $0$ and slope $1$ at the origin: $\sin x$ , $\tan x$ , $\arcsin x$ , $\arctan x$ , $e^{x}-1$ and $\ln(1+x)$ .