AP Calculus AB and BC

Limit of (3^x - 1)/x as x Approaches 0

Q: Why is the base e special?

Because \ln e = 1, so \lim_{x \to 0}\frac{e^{x}-1}{x} = 1 exactly. Every other base picks up a logarithm factor, which is why e is the natural choice for calculus.

Q: Can I use L'Hopital's rule?

Yes, and it gives 3^{x}\ln 3 \to \ln 3. Some instructors object, since this limit is part of how the derivative of 3^{x} is established in the first place.

The limit of three to the x minus one over x as x approaches zero is the natural logarithm of three, about 1.0986. The expression is the limit definition of the derivative of three to the x evaluated at zero.

\lim_{x \to 0} \frac{3^{x}-1}{x} = \ln 3

Settled by the limit definition of the derivative.

Read it as a derivative

Set $f(x) = 3^{x}$ . Since $f(0) = 1$ , the expression is exactly the difference quotient at zero.

\lim_{x \to 0}\frac{3^{x}-1}{x} = \lim_{x \to 0}\frac{f(x)-f(0)}{x-0} = f'(0)

The derivative of $a^{x}$ is $a^{x}\ln a$ , so $f'(0) = 3^{0}\ln 3 = \ln 3$ .

Why the logarithm appears at all

Write $3^{x} = e^{x\ln 3}$ . Then the numerator is $e^{x\ln 3} - 1$ , which behaves like $x \ln 3$ for small $x$ , since $e^{u} - 1 \approx u$ near zero.

Dividing by $x$ leaves $\ln 3$ . The general result is $\lim_{x \to 0}\frac{a^{x}-1}{x} = \ln a$ , and $a = e$ recovers the familiar limit of 1.

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 is the base e special?

Because $\ln e = 1$ , so $\lim_{x \to 0}\frac{e^{x}-1}{x} = 1$ exactly. Every other base picks up a logarithm factor, which is why e is the natural choice for calculus.

Can I use L'Hopital's rule?

Yes, and it gives $3^{x}\ln 3 \to \ln 3$ . Some instructors object, since this limit is part of how the derivative of $3^{x}$ is established in the first place.