L'Hopital's Rule states that if lim f(x)/g(x) produces an indeterminate form 0/0 or infinity/infinity, then the limit equals lim f'(x)/g'(x), provided this latter limit exists. The rule can be applied repeatedly for persistent indeterminate forms. Other indeterminate forms (0 * infinity, infinity - infinity, 0^0, 1^infinity, infinity^0) can be converted to 0/0 or infinity/infinity form first.
Verify indeterminate form before applying. Practice with 0/0 and infinity/infinity cases. Then learn to convert other indeterminate forms. Compare with algebraic techniques (factoring, rationalizing) which sometimes work better. Emphasize that L'Hopital's Rule applies to f'/g', not (f/g)'.
From limits at infinity and derivatives, you know that most limits can be evaluated by substitution — just plug in the limiting value and simplify. The problem arises when substitution produces a form like 0/0 or ∞/∞. These are called indeterminate forms because the expression itself gives no information about the limit's value — the limit could be any number, or it might not exist. For example, lim (sin x)/x as x → 0 gives 0/0, yet the limit is 1. L'Hôpital's Rule resolves this by turning a limit of a ratio into a limit of a ratio of derivatives: if lim f(x)/g(x) is 0/0 or ∞/∞, then lim f(x)/g(x) = lim f′(x)/g′(x), provided the latter limit exists.
The critical procedure is always to check the indeterminate form first. Only 0/0 and ∞/∞ qualify directly. If the form is 3/0, the limit is ±∞ (not indeterminate — you don't need the rule, and applying it would be wrong). Once you confirm the form, differentiate numerator and denominator separately — this is not the quotient rule, which differentiates the whole fraction as a single entity. The rule says replace f/g with f′/g′, not with (f/g)′. This is the most common algebraic error.
Other indeterminate forms — 0·∞, ∞−∞, 0⁰, 1^∞, ∞⁰ — are handled by converting them to 0/0 or ∞/∞ first. For 0·∞: rewrite f·g as f/(1/g) or g/(1/f). For ∞−∞: find a common denominator or multiply by a conjugate. For exponential forms like 1^∞: take the natural log first — ln(f(x)^g(x)) = g(x)·ln(f(x)), which converts the problem to 0·∞ form, then apply L'Hôpital, then exponentiate at the end. The standard example is lim (1 + 1/x)^x as x → ∞: taking the log gives x·ln(1 + 1/x), a 0·∞ form that resolves to 1, so the original limit is e¹ = e.
Algebraic methods — factoring, rationalizing, known limits like sin(x)/x — are often faster and should be preferred when they apply. L'Hôpital's Rule is a fallback, not a first resort. The rule can also be applied repeatedly if the new limit is still indeterminate, but watch for loops: trying to evaluate lim (eˣ/eˣ) by L'Hôpital keeps reproducing eˣ/eˣ = 1, which is the answer all along — just simplify directly. The rule will lead you in circles only when the limit is already accessible by simpler means.