AP Calculus AB and BC

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

Q: Is this limit just the derivative of e^x at x = 0?

Yes, written as a difference quotient rather than as f'(0). Anything of the form \lim_{x \to 0}\frac{f(x)-f(0)}{x} is f'(0), and here f(t) = e^t, whose graph crosses the y-axis with slope 1. Spotting that shape is the fastest way through a great many \frac{0}{0} limits.

Q: What is \lim_{x \to 0} \frac{e^{kx}-1}{x}?

It is k. Multiply and divide by k to rebuild the standard form: \frac{e^{kx}-1}{x} = k \cdot \frac{e^{kx}-1}{kx}, and the second factor tends to 1 as kx \to 0. For example \lim_{x \to 0}\frac{e^{3x}-1}{x} = 3.

Q: Do the left and right sides agree?

They do. For x < 0 the numerator e^x - 1 is negative and so is x, so the quotient stays positive and still climbs to 1. The two-sided limit exists and equals 1, which is what makes the derivative of e^x at 0 exist at all.

The limit of (e^x - 1)/x as x approaches 0 is 1. Direct substitution gives 0/0, so the quotient has to be recognized rather than computed: it is the difference quotient for e^x based at 0, so its limit is the slope of y = e^x where it crosses the y-axis, and that slope is 1 by the definition of e.

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

Settled by the limit definition of the derivative.

Reading the quotient as a difference quotient

Set $f(t) = e^t$ . Then $f(0) = 1$ , so the numerator $e^x - 1$ is exactly $f(x) - f(0)$ , and the whole fraction is the difference quotient for $f$ based at the point $0$ with step size $x$ .

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

A difference quotient with the step shrinking to $0$ is the definition of the derivative at the base point, so the limit is $f'(0)$ . And $f'(0) = 1$ is not a rule imported from elsewhere: it is the property that picks $e$ out of every possible base, the one whose graph crosses the $y$ -axis with slope exactly $1$ .

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

That property is also what makes $e$ a particular number rather than a convenience. The graph of $2^x$ crosses the $y$ -axis with slope about $0.69$ and $3^x$ with slope about $1.10$ , so somewhere between $2$ and $3$ sits the base whose crossing slope is exactly $1$ . That base is $e \approx 2.71828$ , and this limit is the statement of it.

What direct substitution gives

Putting $x = 0$ into the fraction kills both halves at once. On top, $e^0 - 1 = 1 - 1 = 0$ ; on the bottom, the $x$ is $0$ by itself.

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

The form $\frac{0}{0}$ is indeterminate. It carries no information about the answer, only the news that numerator and denominator are both collapsing, so the value depends on which one collapses faster.

The usual repairs do not apply. There is no polynomial factor of $x$ to pull out of $e^x - 1$ , and no radical for a conjugate to clear. What the shape wants is recognition.

L'Hopital's rule reaches the same 1

Since the form is $\frac{0}{0}$ , L'Hopital's rule is available. Differentiating top and bottom separately gives $e^x$ over $1$ , and that is $1$ at $x = 0$ .

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

Fast, but it borrows the answer

L'Hopital's rule needs $\frac{d}{dx}\left[e^x\right] = e^x$ before it can run. In courses that prove that derivative from this very limit, quoting L'Hopital here argues in a circle, while the difference quotient above appeals only to the slope-at- $0$ property that defines $e$ . Both lines earn full credit on the AP exam; only one of them explains anything.

The mistakes students make

Answering $0$ because the numerator goes to $0$ . The denominator goes to $0$ at the same rate, and the ratio settles at $1$ .
Answering that the limit does not exist because $f(0)$ is undefined. A limit describes the approach, and never uses the value at the point itself.
Confusing this with $\lim_{x \to 0} \frac{e^x}{x}$ , whose form is $\frac{1}{0}$ . That one has no finite value: it runs to $+\infty$ from the right and $-\infty$ from the left.
Offering $e^x \approx 1 + x$ as the justification. The approximation is correct near $0$ , but it is the same statement as the limit, so on its own it proves nothing.

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

Is this limit just the derivative of $e^x$ at $x = 0$ ?

Yes, written as a difference quotient rather than as $f'(0)$ . Anything of the form $\lim_{x \to 0}\frac{f(x)-f(0)}{x}$ is $f'(0)$ , and here $f(t) = e^t$ , whose graph crosses the $y$ -axis with slope $1$ . Spotting that shape is the fastest way through a great many $\frac{0}{0}$ limits.

What is $\lim_{x \to 0} \frac{e^{kx}-1}{x}$ ?

It is $k$ . Multiply and divide by $k$ to rebuild the standard form: $\frac{e^{kx}-1}{x} = k \cdot \frac{e^{kx}-1}{kx}$ , and the second factor tends to $1$ as $kx \to 0$ . For example $\lim_{x \to 0}\frac{e^{3x}-1}{x} = 3$ .

Do the left and right sides agree?

They do. For $x < 0$ the numerator $e^x - 1$ is negative and so is $x$ , so the quotient stays positive and still climbs to $1$ . The two-sided limit exists and equals $1$ , which is what makes the derivative of $e^x$ at $0$ exist at all.