AP Calculus AB and BC

Limit of 5^x as x Approaches Negative Infinity

Q: What is the limit of 5^x as x approaches negative infinity?

It is 0, approached from above.

Q: Is this true for any base?

For any base greater than 1, yes. For a base between 0 and 1 the behaviour at the two ends is swapped.

Q: How is 5^x different from x^5?

x^{5} is a power function and tends to -\infty at negative infinity. 5^{x} is an exponential and tends to 0.

The limit of 5 to the x as x approaches negative infinity is 0. A negative exponent makes it 1 over 5 to the positive power, and that denominator grows without bound. The same holds for any base greater than 1.

\lim_{x \to -\infty} 5^{x} = 0

Settled by rewriting as a reciprocal.

What a negative exponent does

Substituting $x = -t$ with $t \to \infty$ makes the behaviour obvious.

5^{-t} = \frac{1}{5^{t}} \longrightarrow 0 \quad \text{as } t \to \infty

The values are always strictly positive, so the graph approaches the horizontal asymptote $y = 0$ from above and never crosses it.

The base decides the direction

For a base greater than 1 the function decays to 0 at minus infinity and grows at plus infinity. For a base between 0 and 1 the two ends swap. Base exactly 1 is the constant function.

The general rule for exponentials

$a > 1$ : $\lim_{x \to -\infty}a^{x} = 0$ and $\lim_{x \to \infty}a^{x} = \infty$
$0 < a < 1$ : $\lim_{x \to -\infty}a^{x} = \infty$ and $\lim_{x \to \infty}a^{x} = 0$
$a = 1$ : constant, so both limits are $1$

Every exponential passes through $(0, 1)$ , since $a^{0} = 1$ for any positive base, which is a useful anchor when sketching.

The mistakes students make

Answering $-\infty$ because the input is going to $-\infty$ . An exponential with a positive base is never negative.
Confusing $5^{x}$ with $x^{5}$ . The power function $x^{5}$ does go to $-\infty$ ; the exponential does not.
Assuming the graph crosses the $x$ -axis. It approaches $y = 0$ asymptotically and stays strictly above it.

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 5^x as x approaches negative infinity?

It is $0$ , approached from above.

Is this true for any base?

For any base greater than $1$ , yes. For a base between $0$ and $1$ the behaviour at the two ends is swapped.

How is 5^x different from x^5?

$x^{5}$ is a power function and tends to $-\infty$ at negative infinity. $5^{x}$ is an exponential and tends to $0$ .