AP Calculus AB and BC

Limit of arctan x / sqrt x at Infinity Is 0

Q: Can I use L'Hopital?

No. The numerator tends to \frac{\pi}{2}, not to infinity, so the form is not one the rule accepts.

The limit of arctan x over the square root of x, as x approaches infinity, is 0. Arctangent flattens toward pi over 2 while the square root grows without bound, so a bounded numerator over an unbounded denominator gives 0.

\lim_{x \to \infty} \frac{\arctan x}{\sqrt{x}} = 0

Settled by bounded over unbounded.

No indeterminate form at all

The numerator tends to $\frac{\pi}{2}$ , a finite number, and the denominator tends to infinity. That is not indeterminate: a finite quantity over an unbounded one is $0$ .

\lim_{x \to \infty}\frac{\arctan x}{\sqrt{x}} = \frac{\pi/2}{\infty} = 0

Check the form before reaching for L'Hopital

The rule needs 0/0 or infinity/infinity. Applying it here is not just unnecessary, it is illegal, and it happens to give the right answer only by accident.

Slow denominators still win

$\sqrt{x}$ grows slowly, but slowly is enough. The limit would still be $0$ with $\ln x$ underneath, since a bounded numerator loses to anything unbounded.

The mistakes students make

Applying L'Hopital's rule. The form is finite over infinite, which the rule does not accept.
Answering $\frac{\pi}{2}$ and ignoring the denominator.

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 / sqrt x at infinity?

It is $0$ .

Can I use L'Hopital?

No. The numerator tends to $\frac{\pi}{2}$ , not to infinity, so the form is not one the rule accepts.