AP Calculus AB and BC

Limit of (2^x-1)/x as x Approaches 0 Is ln 2

Q: Why is the answer \ln 2 instead of 1?

Because 1 is the slope of y = e^x at the y-axis, not the slope of y = 2^x. The graph of 2^x crosses at a shallower angle, slope about 0.6931, and this limit measures precisely that slope. Base e is the single base whose crossing slope is exactly 1.

Q: Does L'Hopital's rule settle it faster?

It does. The form is \frac{0}{0}, and differentiating top and bottom gives \frac{2^x\ln 2}{1}, which is \ln 2 at x = 0. The catch is that it leans on \frac{d}{dx}\left[2^x\right] = 2^x\ln 2, a formula usually proved from this very limit, so as a justification it can run in a circle.

Q: What does the limit become for other bases?

It is \ln a for base a: about 2.3026 for a = 10, exactly 1 for a = e, and 0 for a = 1, where the function is the constant 0 away from the origin. For a < 1 the value is negative, since \ln a < 0 there.

The limit of (2^x - 1)/x as x approaches 0 is the natural log of 2, about 0.6931. Direct substitution gives 0/0. The quotient is the difference quotient for 2^x based at 0, so the limit is the slope of y = 2^x where it crosses the y-axis, and that slope is ln 2 rather than 1.

\lim_{x \to 0} \frac{2^x-1}{x} = \ln 2

Settled by the limit definition of the derivative.

Reading the quotient as a difference quotient

Set $f(t) = 2^t$ . Then $f(0) = 1$ , so the numerator $2^x - 1$ is $f(x) - f(0)$ and the fraction is the difference quotient for $f$ based at $0$ .

\frac{2^x-1}{x} = \frac{f(0+x)-f(0)}{x}, \qquad f(t) = 2^t

The limit is therefore $f'(0)$ , the slope of $y = 2^x$ at its $y$ -intercept. To pin that number down, rewrite the base through $e$ using $2 = e^{\ln 2}$ .

2^x = \left(e^{\ln 2}\right)^x = e^{x\ln 2}

Then force the denominator to match the exponent. Multiplying and dividing by $\ln 2$ rebuilds the one exponential limit that is already known, $\lim_{u \to 0}\frac{e^u-1}{u} = 1$ .

\frac{2^x-1}{x} = \ln 2 \cdot \frac{e^{x\ln 2}-1}{x\ln 2}

With $u = x\ln 2$ , the second factor tends to $1$ as $x \to 0$ , and the constant out front survives untouched.

\lim_{x \to 0} \frac{2^x-1}{x} = \ln 2 \approx 0.6931

What direct substitution gives

At $x = 0$ the numerator is $2^0 - 1 = 1 - 1 = 0$ , and the denominator is $0$ .

\frac{2^0-1}{0} = \frac{0}{0}

Indeterminate, and there is nothing to factor and no radical to rationalize. Sampling close to $0$ shows what the form hides: each base produces its own number, and that number is the logarithm of the base.

$a$	$\frac{a^{0.001}-1}{0.001}$	$\ln a$
$2$	$0.69339$	$0.69315$
$3$	$1.09922$	$1.09861$
$10$	$2.30524$	$2.30259$

The general rule this proves

Nothing in the derivation used the number $2$ , so the same lines work for any base $a > 0$ .

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

That is where the derivative formula for a general exponential comes from. Writing $a^{x+h} = a^x \cdot a^h$ pulls $a^x$ out of the difference quotient and leaves exactly this limit behind.

\frac{d}{dx}\left[a^x\right] = a^x \lim_{h \to 0} \frac{a^h-1}{h} = a^x \ln a

Why e is the convenient base

The stray factor $\ln a$ equals $1$ exactly when $a = e$ , which is what makes $\frac{d}{dx}\left[e^x\right] = e^x$ and leaves every other base carrying a constant. For $a = 2$ that constant is about $0.6931$ , so $y = 2^x$ rises a little more gently than its own height.

The mistakes students make

Answering $1$ by pattern matching against $\lim_{x \to 0}\frac{e^x-1}{x}$ . That one is $1$ because its base is $e$ , and changing the base changes the answer.
Answering $2$ , or leaving a variable in the answer such as $x\ln 2$ . The limit is a single number, $\ln 2 \approx 0.6931$ .
Differentiating $2^x$ with the power rule as $x \cdot 2^{x-1}$ . The exponent is the variable here, so the power rule does not apply and the derivative is $2^x\ln 2$ .
Treating $\frac{0}{0}$ as $1$ because the two parts hit zero together. Arriving together says nothing about the ratio on the way in.

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 answer $\ln 2$ instead of $1$ ?

Because $1$ is the slope of $y = e^x$ at the $y$ -axis, not the slope of $y = 2^x$ . The graph of $2^x$ crosses at a shallower angle, slope about $0.6931$ , and this limit measures precisely that slope. Base $e$ is the single base whose crossing slope is exactly $1$ .

Does L'Hopital's rule settle it faster?

It does. The form is $\frac{0}{0}$ , and differentiating top and bottom gives $\frac{2^x\ln 2}{1}$ , which is $\ln 2$ at $x = 0$ . The catch is that it leans on $\frac{d}{dx}\left[2^x\right] = 2^x\ln 2$ , a formula usually proved from this very limit, so as a justification it can run in a circle.

What does the limit become for other bases?

It is $\ln a$ for base $a$ : about $2.3026$ for $a = 10$ , exactly $1$ for $a = e$ , and $0$ for $a = 1$ , where the function is the constant $0$ away from the origin. For $a < 1$ the value is negative, since $\ln a < 0$ there.