AP Calculus AB and BC

Limit of e^(-x) as x Approaches Infinity Is 0

Q: What is the limit of e^(-x) as x approaches infinity?

It is 0, approached from above since the function is always positive.

Q: What is the limit as x approaches negative infinity?

It is \infty: the exponent -x becomes large and positive, so the function grows without bound.

Q: Does x^n e^(-x) also go to 0?

Yes, for every fixed n. Exponential decay beats polynomial growth.

The limit of e to the minus x as x approaches infinity is 0. Writing it as 1 over e to the x makes it clear: the denominator grows without bound, so the fraction shrinks to nothing. The values stay strictly positive, so the approach is from above.

\lim_{x \to \infty} e^{-x} = 0

Settled by rewriting as a reciprocal.

Reading it as a reciprocal

A negative exponent is a reciprocal, and that turns the question into one about growth.

e^{-x} = \frac{1}{e^{x}}

As $x \to \infty$ the denominator $e^{x}$ grows without bound, so the fraction goes to $0$ . It never reaches $0$ , because an exponential is positive for every real input.

\lim_{x \to \infty}e^{-x} = 0

The other end is the opposite

As x goes to minus infinity, e^(-x) grows without bound, because the exponent minus x becomes large and positive. The graph is the mirror image of e^x.

Why exponential decay beats every power

This limit is why $x^{n}e^{-x} \to 0$ for every fixed power $n$ . The exponential shrinks faster than any polynomial grows, which is the growth ordering BC students are expected to know.

\lim_{x \to \infty}x^{n}e^{-x} = 0 \quad \text{for every } n

In context, $e^{-x}$ is the shape of every decay model in Unit 7: a quantity falling at a rate proportional to how much remains, approaching zero without a finite time at which it arrives.

The mistakes students make

Answering that the function reaches $0$ . It approaches $0$ as a horizontal asymptote and is positive everywhere.
Reading $e^{-x}$ as $-e^{x}$ . A negative exponent is a reciprocal, not a negative value; $e^{-x}$ is always positive.
Assuming a polynomial factor can rescue it. Even $x^{100}e^{-x}$ tends to $0$ , because exponentials outpace every power.

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 e^(-x) as x approaches infinity?

It is $0$ , approached from above since the function is always positive.

What is the limit as x approaches negative infinity?

It is $\infty$ : the exponent $-x$ becomes large and positive, so the function grows without bound.

Does x^n e^(-x) also go to 0?

Yes, for every fixed $n$ . Exponential decay beats polynomial growth.