L'Hôpital's RuleJEE Main

Concept

When a limit lands on 0/0 or ∞/∞, the ratio's fate is decided by how FAST top and bottom vanish (or blow up) — i.e. by their derivatives: $\lim \frac{f}{g} = \lim \frac{f'}{g'}$ when the right side exists. It's Taylor's first-order terms fighting it out.

Key formula

Derivation

Near a: $f(x) \approx f'(a)(x-a)$ and $g(x) \approx g'(a)(x-a)$ (both f(a) = g(a) = 0). The (x−a) cancels, leaving f'(a)/g'(a).

If still 0/0 — apply again ((1−cosx)/x² needs one more round or the sinx/2x shortcut). Other indeterminate forms (0·∞, ∞−∞, 1^∞, 0⁰) must be ALGEBRAICALLY rearranged into a quotient first.

Scenarios to explore

• L'Hôpital's Rule — 0/0 forms cracked by differentiating top and bottom.

Real-world applications

• The standard-limit factory: sinx/x, (eˣ−1)/x, ln(1+x)/x — all equal 1.
• Asymptotic comparisons: ln x loses to every power, powers lose to eˣ.
• 1^∞ forms via exp-log: lim (1 + 1/x)ˣ = e.

JEE exam tips

• Check the form FIRST — 0/0 or ∞/∞ only.
• Series expansion often beats repeated L'Hôpital: 1 − cosx ≈ x²/2 kills the second example instantly.
• 1^∞: rewrite as e^{lim g·(f−1)} — the standard exponential bypass.

Common mistakes

• Applying it to NON-indeterminate forms (lim x/(x+1) at 0 is plainly 0 — L'Hôpital gives garbage… actually 1: wrong!).
• Quotient-rule instead of separate derivatives (it's f'/g', NOT (f/g)').
• Circular reasoning: proving sinx/x = 1 by L'Hôpital uses the derivative of sin, which uses that very limit.

Exam traps to avoid

• If f'/g' oscillates without a limit, L'Hôpital is silent — the original limit may still exist.
• x/(x + sinx) as x→∞: L'Hôpital cycles forever; divide by x instead → 1.