AP Calculus AB and BC

Limit of (e^x - 1)/sin x at 0 Is 1

Q: Do I need L'Hopital's rule here?

No. Dividing through by x splits the quotient into \frac{e^{x}-1}{x} over \frac{\sin x}{x}, and both tend to 1. L'Hopital reaches the same 1 if you prefer it.

Q: Why does e^x - 1 behave like x near 0?

Because e^{x} leaves the point (0,1) with slope 1. Subtracting the 1 leaves a curve that hugs the line y = x close to the origin.

The limit of e to the x minus 1, over sine of x, as x approaches 0 is 1. Divide the numerator and the denominator by x. That turns the quotient into one standard limit over another, and each of those tends to 1, so the ratio is 1.

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

Settled by splitting into two standard limits.

Divide top and bottom by x

\frac{e^{x}-1}{\sin x} = \frac{\dfrac{e^{x}-1}{x}}{\dfrac{\sin x}{x}}

Nothing has been changed except the bookkeeping, and now both halves are limits worth knowing on sight: $\lim_{x \to 0}\frac{e^{x}-1}{x} = 1$ and $\lim_{x \to 0}\frac{\sin x}{x} = 1$ .

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

Why both pieces equal 1

Each is a derivative in disguise. The first is the derivative of e^x at 0, the second is the derivative of sin x at 0, and both of those derivatives equal 1. That is the whole reason the two famous limits agree.

L'Hopital gives the same answer

Substitution produces $\frac{0}{0}$ , so differentiating the top and the bottom separately is allowed.

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

One line either way. The split version is worth practising because it works when a coefficient sits inside the sine and the derivatives get messier.

The mistakes students make

Every one of these comes from misreading the numerator or from treating an indeterminate form as a value.

Answering $0$ because $e^{0}-1 = 0$ . The denominator vanishes as well, so the form is $\frac{0}{0}$ and decides nothing.
Reading $e^{x}-1$ as $e^{x-1}$ . That is $\frac{1}{e}$ at $x = 0$ over a vanishing denominator, so the quotient looks infinite.
Answering $e$ by approximating $e^{x}-1$ as $ex$ . The correct linear approximation near $0$ is $e^{x}-1 \approx x$ , with no factor of $e$ .

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 (e^x - 1)/sin x as x approaches 0?

It is $1$ .

Do I need L'Hopital's rule here?

No. Dividing through by $x$ splits the quotient into $\frac{e^{x}-1}{x}$ over $\frac{\sin x}{x}$ , and both tend to $1$ . L'Hopital reaches the same $1$ if you prefer it.

Why does e^x - 1 behave like x near 0?

Because $e^{x}$ leaves the point $(0,1)$ with slope $1$ . Subtracting the $1$ leaves a curve that hugs the line $y = x$ close to the origin.