AP Calculus AB and BC

Limit of x^2/e^x at Infinity Is 0

The limit of x squared over e to the x as x approaches infinity is 0. Two passes of L'Hopital's rule reduce the numerator to a constant while the denominator stays exponential. Exponential growth beats every polynomial, no matter how high the power.

lim⁡x→∞x2ex=0\lim_{x \to \infty} \frac{x^{2}}{e^{x}} = 0

Settled by L'Hopital's rule twice.

Differentiate the polynomial away

lim⁡x→∞x2ex  =H  lim⁡x→∞2xex  =H  lim⁡x→∞2ex=0\lim_{x \to \infty}\frac{x^{2}}{e^{x}} \;\overset{\text{H}}{=}\; \lim_{x \to \infty}\frac{2x}{e^{x}} \;\overset{\text{H}}{=}\; \lim_{x \to \infty}\frac{2}{e^{x}} = 0

Each pass lowers the power by one while the exponential is unchanged. After nn passes the numerator is a constant, so xnex→0\frac{x^{n}}{e^{x}} \to 0 for every fixed nn.

The growth ordering in one fact

ln x, then any power of x, then any exponential, then n factorial, then n^n. Each is beaten by the next, and this limit is the third link.

The mistakes students make

  • Answering ∞\infty because the numerator grows. It does, just far too slowly.
  • Stopping after one pass at 2xex\frac{2x}{e^{x}}, which is still ∞∞\frac{\infty}{\infty}.
  • Assuming a big enough power wins. Even x100x^{100} loses.

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 x^2/e^x at infinity?

It is 00.

Does a higher power change it?

No. xnex→0\frac{x^{n}}{e^{x}} \to 0 for every fixed nn; it just takes nn passes of L'Hopital.