AP Calculus AB and BC

Limit of x^3/e^x at Infinity Is 0

The limit of x cubed over e to the x as x approaches infinity is 0. Three passes of L'Hopital's rule take the numerator down to a constant while the denominator remains e to the x. The number of passes equals the power.

\lim_{x \to \infty} \frac{x^{3}}{e^{x}} = 0

Settled by L'Hopital's rule three times.

Three passes, one per degree

\frac{x^{3}}{e^{x}} \;\to\; \frac{3x^{2}}{e^{x}} \;\to\; \frac{6x}{e^{x}} \;\to\; \frac{6}{e^{x}} \longrightarrow 0

Every pass is legal because each stage is still $\frac{\infty}{\infty}$ . Checking that before each application is the part worth writing down.

The peak before the fall

The function is not decreasing everywhere. Differentiating gives $\frac{x^{2}(3-x)}{e^{x}}$ , so it rises until $x = 3$ , peaks at $\frac{27}{e^{3}} \approx 1.34$ , and only then decays. A limit at infinity says nothing about the journey.

The mistakes students make

Stopping early while the form is still indeterminate.
Concluding the function decreases everywhere because the limit is $0$ .

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^3/e^x at infinity?

It is $0$ .

How many passes of L'Hopital are needed?

Three, one per degree of the numerator.