Guides / Limits

L'Hôpital's rule: check the form, then differentiate top and bottom separately

L'Hôpital's rule turns a limit that plugs in as 0/0 or ∞/∞ into a limit of derivatives, which is usually easier. It's a fast tool, and almost every mistake with it comes from one of two places: using it on a limit that isn't indeterminate, or differentiating the fraction as a whole instead of the top and bottom separately.

Why it works

Near a point a where f(a) = g(a) = 0, both functions look like their tangent lines: f(x) ≈ f′(a)(x − a) and g(x) ≈ g′(a)(x − a). The factor (x − a) cancels, and the ratio of the functions approaches the ratio of their slopes, f′(a)/g′(a). That's the whole idea: when both the top and the bottom head to zero, what matters is how fast each one gets there.

The same reasoning fails when the top and bottom don't both go to 0 (or both to ∞). Then the limit is already decided by plugging in, and the ratio of the slopes is a different number with nothing to do with it.

Example 1: 0/0

Find lim_x→0 (e^2x − 1)/x.

1. (e⁰ − 1)/0 = 0/0Plug in first. 0/0 is indeterminate, so the rule applies.
2. lim_x→0 2e^2x1Top: the derivative of e²ˣ − 1 is 2e²ˣ (chain rule). Bottom: the derivative of x is 1.
3. = 2e⁰ = 2This limit isn't indeterminate, so plugging in finishes it.

Example 2: applying the rule more than once

Find lim_x→0 (x − sin x)/x³.

1. (0 − 0)/0 = 0/0Indeterminate.
2. lim_x→0 1 − cos x3x²Differentiate top and bottom. Plugging in gives 0/0 again.
3. lim_x→0 sin x6xApply the rule a second time. Still 0/0.
4. lim_x→0 cos x6A third time. Now the bottom is 6, not 0.
5. = 16cos 0 = 1. Check the form before every application, not just the first.

Example 3: ∞/∞

Find lim_x→∞ (3x² + x)/(5x² − 2).

1. ∞/∞Both the top and the bottom grow without bound.
2. lim_x→∞ 6x + 110xDifferentiate top and bottom. Still ∞/∞.
3. lim_x→∞ 610Once more.
4. = 35The same answer as dividing top and bottom by x², the method in the limits guide.

Example 4: 0 · ∞, rewritten as a fraction

Find lim_x→0⁺ x² ln x.

1. 0 · (−∞)A product, not a fraction, so the rule can't be used yet.
2. x² ln x = ln xx⁻²Move one factor to the bottom as its reciprocal. Now it's −∞/∞.
3. lim 1/x−2x⁻³ = lim −x²2Differentiate top and bottom, then simplify: (1/x) · (x³/−2).
4. = 0Put the log on top. Moving x² down as 1/ln x would give a harder derivative.

Example 5: 1^∞, using a log

Find lim_x→0 (1 + 3x)^1/x.

1. 1^∞The base goes to 1 and the exponent to ∞. That's indeterminate, not 1.
2. ln L = lim_x→0 ln(1 + 3x)xTake the log of the expression. The power comes down: ln(a^b) = b ln a. Now it's 0/0.
3. = lim_x→0 3/(1 + 3x)1 = 3The rule, with the chain rule on top.
4. L = e³ ≈ 20.09Step 3 found the limit of the log. Undo it: the limit itself is e³.

Common mistakes, and the exact line they happen on

1. Using the rule when the limit isn't indeterminate

Find lim_x→1 (x² + 1)/(x + 1).

1. lim_x→1 2x1 = 2This is the mistake. Plugging in gives 2/2, which isn't 0/0 or ∞/∞, so the rule says nothing here. The limit is just 2/2 = 1. Always plug in before differentiating.

2. Using the quotient rule

Find lim_x→0 (sin x)/x.

1. 0/0Right so far.
2. lim_x→0 x cos x − sin xx² = 0This is the mistake. That's the derivative of the whole fraction (the quotient rule), which is a different function with a different limit. L'Hôpital's rule differentiates the top and bottom on their own: cos x / 1 → 1.

3. Reading 1^∞ as 1, or forgetting to undo the log

Find lim_x→0 (1 + 3x)^1/x (Example 5).

1. 1 + 3x → 1, and 1 to any power is 1, so the limit is 1This is the mistake. The base is never exactly 1, and the exponent keeps growing, so the two effects compete. The limit is e³ ≈ 20.09.
2. lim ln(1 + 3x)/x = 3, so the limit is 3The other version. 3 is the limit of the log. The answer is e³, not 3. Write ln L = … at the start so the last step can't get lost.

Practice

1. Find lim_x→0 (tan 4x)/x.

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   0/0. Differentiate: 4 sec²(4x)/1 → 4.

2. Find lim_x→1 (ln x)/(x − 1).

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   0/0. Differentiate: (1/x)/1 → 1.

3. Find lim_x→0 (1 − cos 2x)/x².

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   0/0, then 2 sin 2x / (2x), still 0/0, then 4 cos 2x / 2 → 2.

4. Find lim_x→∞ x³/e^x.

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   ∞/∞ three times: 3x²/e^x, 6x/e^x, 6/e^x → 0. Exponentials beat powers.

5. Find lim_x→∞ x^1/x.

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   ∞⁰. Take a log: ln L = lim (ln x)/x, which is ∞/∞ and gives (1/x)/1 → 0. So L = e⁰ = 1.

6. Find lim_x→0 (e^x − 1)/sin x.

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   0/0. Differentiate: e^x/cos x → 1/1 = 1.