L' Hopital Rule

L' Hopital Rule

1. Statement

If f(x) and g(x) are differentiable near x = a (a may be finite or infinite), with g′(x) ≠ 0 near a, and the quotient f(x)/g(x) takes the indeterminate form 0/0 or ∞/∞ as x → a, then

provided the limit on the right exists (it may be finite, +∞ or −∞). The rule also holds for one-sided limits and for x → ±∞.

Crucial point: we differentiate the numerator and denominator separately — this is not the quotient rule (which differentiates the whole fraction).

2. Conditions and Cautions

1. Check the form first. Only 0/0 and ∞/∞ may be attacked directly. A 0 × ∞, ∞ − ∞, 0⁰, 1^∞ or ∞⁰ form must first be converted (factor into a quotient, or take logarithms — see the indeterminate forms topic).
2. Re-check after each application. The new quotient may still be 0/0 or ∞/∞; if so, apply the rule again. Stop as soon as the form becomes determinate.
3. The rule can fail if the new limit does not exist. Classic counter-example:

        lim x→∞ (x + sin x)/x           →  ∞/∞ form; differentiate
        (1 + cos x)/1  does NOT have a limit as x → ∞ — yet the original limit EXISTS:
        (x + sin x)/x = 1 + (sin x)/x → 1 + 0 = 1
   ⇒  the failure of the derived limit tells us nothing; a direct simplification wins here.
   

4. Sometimes algebra or a series expansion is better. For (x² − 1)/(x − 1) or sin 3x/sin 5x, factorising or standard limits are one line, whereas L'Hôpital is two.
5. The rule is valid for 0/0 and ∞/∞ only — using it on, say, 1^∞ by "differentiating the powers" is a common mistake. Convert to a quotient first.

3. Proof Outline (using Cauchy's Mean Value Theorem)

For the 0/0 case with f(a) = g(a) = 0, apply Cauchy's MVT on the interval [a, x]:

Since f(a) = g(a) = 0, the left side is f(x)/g(x); and as x → a the intermediate point c → a too. Hence lim f/g = lim f′/g′. The ∞/∞ case is handled by a similar argument applied to 1/f and 1/g.

   (a)  lim x→0  (sin x)/x            [0/0]
        = lim (cos x)/1 = cos 0 = 1                                 ✓

   (b)  lim x→0  (1 - cos x)/x²       [0/0]
        = lim (sin x)/(2x)            [still 0/0 — differentiate again]
        = lim (cos x)/2 = 1/2                                       ✓

   (c)  lim x→0  (e^x - 1 - x)/x²     [0/0]
        = lim (e^x - 1)/(2x)          [still 0/0]
        = lim (e^x)/2 = 1/2                                        ✓

   (d)  lim x→∞  x e^(−x) = lim x/e^x   [∞/∞]
        = lim 1/e^x = 0                                            ✓

   (e)  lim x→∞  (log x)/x            [∞/∞]
        = lim (1/x)/1 = 0                                          ✓

   (f)  lim x→0  (tan x - x)/x³       [0/0 — three applications]
        = lim (sec²x - 1)/(3x²)
        = lim (2 sec²x tan x)/(6x)     [using d/dx sec²x = 2 sec²x tan x]
        = (1/3) lim (sec²x)(tan x/x) = (1/3)(1)(1) = 1/3            ✓

6. When NOT to Use L'Hôpital

Limit	Better method
lim x→2 (x² − 4)/(x − 2)	factorise → 4
lim x→∞ (√(x² + x) − x)	rationalise → 1/2
lim x→0 sin 3x / sin 5x	standard limit → 3/5
lim x→∞ (x + sin x)/x	divide by x → 1 (L'Hôpital fails)
lim x→0 (1 + x)^(1/x)	standard limit → e

7. Comparison with the Series Method

For 0/0 forms involving standard functions, expanding in powers often finishes in one step:

   lim x→0 (x - sin x)/x³ = lim [x - (x - x³/3! + x⁵/5! - ...)]/x³ = lim [x³/6 - ...]/x³ = 1/6
   (the same answer L'Hôpital gives after three differentiations)

8. Exam Points

• State L'Hôpital's rule with its conditions — 3 marks, almost certain.
• Evaluate limits of 0/0 and ∞/∞ type using the rule (including repeated application) — 5-7 marks, almost certain.
• Evaluate limits requiring conversion first: ∞ − ∞ (sec x − tan x), 0 × ∞ (x log x), 1^∞ ((1 + a/x)^x), 0⁰ (x^x) — 5-7 marks, very frequent.
• Prove the rule using Cauchy's Mean Value Theorem — 5 marks.
• Show by counter-example that the rule can fail when the derived limit does not exist — 3-5 marks, frequent.
• Explain why numerator and denominator are differentiated separately (not by the quotient rule) — 3 marks.