AP Calculus AB and BC

Limit of e^(-x^2) as x Approaches Infinity

The limit of e to the negative x squared as x approaches infinity is zero. The exponent runs to negative infinity, and it does so quadratically, which is why this bell-shaped curve decays faster than any ordinary exponential.

lim⁡x→∞e−x2=0\lim_{x \to \infty} e^{-x^{2}} = 0

Settled by end behaviour of the exponential.

Track the exponent

As x→∞x \to \infty, the exponent −x2→−∞-x^{2} \to -\infty, and eu→0e^{u} \to 0 as u→−∞u \to -\infty. So the limit is 0, and y=0y = 0 is a horizontal asymptote.

The decay is dramatic: at x=5x = 5 the value is about 1.4×10−111.4 \times 10^{-11}, and at x=10x = 10 it is around 3.7×10−443.7 \times 10^{-44}. In double-precision arithmetic it underflows to exactly 0 well before x=30x = 30.

The curve this describes

This is the Gaussian, the shape of the normal distribution. Its total area is π\sqrt{\pi}, a striking result given that the function has no elementary antiderivative at all.

That last point is worth remembering: ∫e−x2dx\int e^{-x^{2}}dx cannot be written in closed form, which is why the error function exists as a named function and why AP questions about it are always definite integrals evaluated numerically.

This is the reason AP Statistics hands out z-tables. Since e−x2e^{-x^{2}} has no elementary antiderivative, there is no formula to evaluate the area under a normal distribution between two bounds, so the areas have to be tabulated or computed numerically instead of integrated.

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

Why does this decay faster than e to the minus x?

Because the exponent grows quadratically rather than linearly. By x=10x = 10 the exponent is −100-100 instead of −10-10, a difference of forty orders of magnitude in the value.

Does e to the minus x squared have an antiderivative?

Not an elementary one. The antiderivative is defined as the error function, and definite integrals of it are computed numerically or by the known total area.