Evaluate lim(x→0) (e^x - 1) / x.

Step 1 — Substitute and check the form

At x = 0 the numerator is e^0 − 1 = 0 and the denominator is 0, so the expression is the indeterminate form 0/0. That is the precondition for L'Hopital's rule — checking it first is part of the expected working.

Step 2 — Differentiate numerator and denominator separately

This is not the quotient rule. Each part is differentiated on its own:

• numerator: d/dx (e^x − 1) = e^x
• denominator: d/dx (x) = 1

Step 3 — Take the limit again

lim(x→0) e^x / 1 = e^0 = 1

Alternative without L'Hopital

The same result follows from the definition of the derivative of e^x at x = 0, or from the Maclaurin series e^x = 1 + x + x²/2 + …, where (e^x − 1)/x = 1 + x/2 + … tends to 1.