AP Calculus AB and BC

Limit of e^x as x Approaches Negative Infinity

Q: Does e^x ever actually equal 0?

No. The range of e^{x} is every positive number, so the graph gets arbitrarily close to the x-axis without touching it. That is what a horizontal asymptote means: approached, never attained. It is also why \ln 0 is undefined, since no exponent produces an output of 0.

Q: Where does this limit get used on the AP exam?

It settles horizontal asymptotes for anything built from e^{x}, and it drives exponential decay models: a solution y = Ce^{kt} with k < 0 approaches 0 as t grows because the exponent heads to -\infty. It also makes improper integrals such as \int_{-\infty}^{0} e^{x}\,dx converge, to 1 in that case.

Q: Does the same answer hold for other bases?

For any base a > 1 the behavior matches e^{x} exactly, so \lim_{x \to -\infty} a^{x} = 0. For 0 < a < 1 it flips, because a^{x} = \left(\frac{1}{a}\right)^{-x} grows without bound on the left. The base 1 case is constant.

The limit of e to the x as x approaches negative infinity is 0. The exponential decays toward 0 on the left, which makes y = 0 a horizontal asymptote there, while on the right it grows without bound. The value never reaches 0: e to the x is positive for every real number x.

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

Settled by end behaviour of the exponential.

Turning the negative exponent into a reciprocal

Substituting $x = -t$ converts the awkward direction into a familiar one: as $x \to -\infty$ , the new variable satisfies $t \to \infty$ .

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

Now the behavior is obvious. The denominator $e^{t}$ grows without bound, and a fixed numerator over an unbounded denominator collapses to 0.

\lim_{t \to \infty} \frac{1}{e^{t}} = 0

The numbers show how quickly it happens. Every output is positive and each step of 5 in the input divides the value by roughly 148.

$x$	$e^{x}$
$-1$	$0.3679$
$-5$	$0.006738$
$-10$	$0.0000454$
$-20$	$0.00000000206$

The values are shrinking toward 0 from above and never cross it, which is exactly what a horizontal asymptote at $y = 0$ looks like from the left.

Why substitution fails, and why nothing is indeterminate

Direct substitution is not available, for the same reason it is never available at an infinite endpoint: $-\infty$ is a description of behavior, not a number to plug in. Writing $e^{-\infty}$ is shorthand at best.

What separates this limit from most is that no indeterminate form shows up. There is no $\frac{\infty}{\infty}$ or $\infty - \infty$ to resolve, because nothing is competing: one function, one direction, one shape. The work is knowing the exponential's end behavior rather than doing algebra.

Do not reach for L'Hopital here

L'Hopital's rule needs a quotient in $\frac{0}{0}$ or $\frac{\infty}{\infty}$ form. $e^{x}$ is neither, and differentiating it returns $e^{x}$ forever, so the rule would loop without ever producing an answer.

The mistake students make

The frequent error is answering $-\infty$ , on the reasoning that a very negative input should give a very negative output. Exponentials do not work that way: the output of $e^{x}$ is positive for every real $x$ , so a negative exponent makes the value small, never negative.

The second error is direction bookkeeping. Four related limits get mixed up constantly, and the only way through is to ask whether the exponent is heading up or down.

Limit	Exponent heads to	Value
$\lim_{x \to -\infty} e^{x}$	$-\infty$	$0$
$\lim_{x \to \infty} e^{x}$	$\infty$	$\infty$
$\lim_{x \to -\infty} e^{-x}$	$\infty$	$\infty$
$\lim_{x \to \infty} e^{-x}$	$-\infty$	$0$

Read the exponent first, then apply the single rule: exponent to $-\infty$ gives 0, exponent to $\infty$ gives $\infty$ .

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

Does e^x ever actually equal 0?

No. The range of $e^{x}$ is every positive number, so the graph gets arbitrarily close to the $x$ -axis without touching it. That is what a horizontal asymptote means: approached, never attained. It is also why $\ln 0$ is undefined, since no exponent produces an output of 0.

Where does this limit get used on the AP exam?

It settles horizontal asymptotes for anything built from $e^{x}$ , and it drives exponential decay models: a solution $y = Ce^{kt}$ with $k < 0$ approaches 0 as $t$ grows because the exponent heads to $-\infty$ . It also makes improper integrals such as $\int_{-\infty}^{0} e^{x}\,dx$ converge, to 1 in that case.

Does the same answer hold for other bases?

For any base $a > 1$ the behavior matches $e^{x}$ exactly, so $\lim_{x \to -\infty} a^{x} = 0$ . For $0 < a < 1$ it flips, because $a^{x} = \left(\frac{1}{a}\right)^{-x}$ grows without bound on the left. The base 1 case is constant.