AP Calculus AB and BC

Limit of sqrt x as x Approaches Infinity Is Infinity

The limit of the square root of x as x approaches infinity is infinity. The square root grows without bound, though much more slowly than x: to make the output reach 1000 the input has to reach a million. Slow growth is still unbounded growth.

lim⁡x→∞x=∞\lim_{x \to \infty} \sqrt{x} = \infty

Settled by unbounded growth of a power function.

Slow growth is still unbounded

For any target MM you name, taking x>M2x > M^{2} makes x>M\sqrt{x} > M. Since no bound can hold it, the function grows without bound.

lim⁡x→∞x=∞\lim_{x \to \infty}\sqrt{x} = \infty

The slowness is real but irrelevant to the limit. The derivative 12x\frac{1}{2\sqrt{x}} tends to 00, so the graph flattens forever, and yet it never levels off at a finite height.

Flattening is not the same as leveling off

A graph whose slope tends to 0 can still climb forever. Both sqrt x and ln x do exactly that. A horizontal asymptote requires the VALUES to settle, not the slopes.

Where it sits in the growth ordering

The standard ordering at infinity, from slowest to fastest, is worth carrying as one fact.

ln⁡x  ≪  x  ≪  x  ≪  x2  ≪  ex\ln x \;\ll\; \sqrt{x} \;\ll\; x \;\ll\; x^{2} \;\ll\; e^{x}

Each item is beaten by the next in the sense that the ratio of the slower to the faster tends to 00. That is why xx→0\frac{\sqrt{x}}{x} \to 0 while xln⁡x→∞\frac{\sqrt{x}}{\ln x} \to \infty.

The mistakes students make

  • Answering that the limit is finite because the graph looks flat. Flatness is about the slope; the height still increases forever.
  • Confusing x\sqrt{x} with 1x\frac{1}{\sqrt{x}}, which does tend to 00.
  • Writing that the limit equals infinity and treating that as a number. It is a description of how the limit fails to exist.

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 sqrt x as x approaches infinity?

It is ∞\infty; the function grows without bound.

Does sqrt x have a horizontal asymptote?

No. Its slope tends to 00, but its values keep increasing, and an asymptote needs the values to settle.

How does sqrt x compare with ln x?

x\sqrt{x} grows faster: ln⁡xx→0\frac{\ln x}{\sqrt{x}} \to 0 as x→∞x \to \infty.