Lesson 19: Linear Approximation, Differentials & L'Hôpital's Rule

Opening Question

You know that √25 = 5 exactly. Your calculator is in your backpack, and a problem asks you to estimate √25.3 quickly, by hand.

Here's the idea you'll formalize today: near x = 25, the graph of f(x) = √x is almost indistinguishable from its tangent line. So instead of evaluating the curve, evaluate the line.

The tangent line at x = 25 has slope f'(25). Since f'(x) = 1/(2√x), we get f'(25) = 1/10. The tangent line is

L(x) = 5 + (1/10)(x − 25)

So L(25.3) = 5 + (1/10)(0.3) = 5.03.

The true value is √25.3 ≈ 5.0299. Our estimate is off by about 0.0001 — excellent for a one-line calculation.

Two questions to carry into the lesson: Is 5.03 an overestimate or an underestimate? And how could we have predicted that before checking the true value? The answer is hiding in the shape of the curve.

Core Concepts

Linear (tangent-line) approximation

Near a point x = a, a differentiable function is well approximated by its tangent line. The linearization of f at a is

L(x) = f(a) + f'(a)(x − a)

We then use f(x) ≈ L(x) for x near a. The whole game is choosing an a close to your target where f(a) and f'(a) are easy to compute exactly (a perfect square, x = 0, x = 1, a multiple of π, etc.).

Worked linearization. Estimate ln(1.1) without a calculator.

Choose a = 1 because ln(1) = 0 is exact. With f(x) = ln x, f'(x) = 1/x, so f'(1) = 1. Then

L(x) = ln(1) + 1·(x − 1) = x − 1

so ln(1.1) ≈ L(1.1) = 0.1. (True value: 0.0953….)

Overestimate or underestimate? Use concavity. A tangent line lies on one definite side of the curve:

• If f is concave down near a (f''< 0), the curve bends below its tangent line, so the tangent line sits above the curve and the estimate is an overestimate.
• If f is concave up near a (f''> 0), the curve bends above its tangent line, so the tangent line sits below the curve and the estimate is an underestimate.

For f(x) = ln x, f''(x) = −1/x² < 0, so f is concave down. The graph confirms the tangent line riding above the curve, so L(1.1) = 0.1 is an overestimate of ln(1.1) = 0.0953. ✓

[GRAPH: f(x) = ln(x) with tangent line L(x) = x − 1 at x = 1, window [0, 2.5] × [−1.5, 1]
- Curve passes through (1, 0), increasing, concave down everywhere (bends downward)
- Tangent line L(x) = x − 1 touches the curve at (1, 0)
- Tangent line lies ABOVE the curve for all x ≠ 1
- Highlight x = 1.1: point on line at (1.1, 0.1) sits above point on curve at (1.1, 0.0953)
- Label: "concave down ⇒ tangent above curve ⇒ overestimate"]

The same logic explains the opening problem: f(x) = √x has f''(x) = −1/(4x^{3/2}) < 0, so it's concave down and 5.03 is an overestimate — matching the true value 5.0299.

Differentials

The differential repackages the same tangent-line idea as a statement about change. If y = f(x), define

dy = f'(x) dx

Here dx is a small change in x (you choose it), and dy is the corresponding change along the tangent line. It approximates the actual change Δy = f(x + dx) − f(x):

Δy ≈ dy = f'(x) dx

Differentials shine for error/change estimation. Suppose a cube has side s = 5 cm, measured with possible error ds = 0.1 cm. With volume V = s³,

dV = 3s² ds = 3(5)²(0.1) = 7.5 cm³

So a 0.1 cm error in the side produces roughly a 7.5 cm³ error in the computed volume. (Actual: 5.1³ − 5³ = 7.651; the differential gives a clean, close estimate.)

L'Hôpital's Rule

When you substitute and get an indeterminate form — a meaningless symbol like 0/0 or ∞/∞ whose value isn't determined by the form alone — limits of quotients can be resolved by differentiating top and bottom separately.

L'Hôpital's Rule. Suppose f and g are differentiable on an open interval around a (except possibly at a), and g'(x) ≠ 0 near a. If

lim_{x→a} f(x)/g(x)   is of the form  0/0  or  ±∞/±∞,

then

lim_{x→a} f(x)/g(x) = lim_{x→a} f'(x)/g'(x)

provided the right-hand limit exists (or is ±∞). The rule also holds for one-sided limits and for a = ±∞.

Read the conditions carefully: 1. The form must be 0/0 or ±∞/±∞. If substitution gives 2/0, 0/5, or 3/7, the rule does not apply. 2. You differentiate f and g separately — this is not the quotient rule.

Worked 0/0. Find lim_{x→0} sin(3x)/x. Substituting gives sin(0)/0 = 0/0 — indeterminate. ✓ Apply L'Hôpital:

lim_{x→0} sin(3x)/x = lim_{x→0} 3cos(3x)/1 = 3cos(0) = 3

Worked ∞/∞. Find lim_{x→∞} (ln x)/x. As x → ∞, this is ∞/∞ — indeterminate. ✓ Apply L'Hôpital:

lim_{x→∞} (ln x)/x = lim_{x→∞} (1/x)/1 = lim_{x→∞} 1/x = 0

The numerator grows, but the denominator grows faster, so the ratio collapses to 0.

Sometimes you apply it twice. If the first application still gives 0/0 or ∞/∞, and the hypotheses still hold, apply the rule again. lim_{x→0} (eˣ − 1 − x)/x² gives 0/0; differentiating gives (eˣ − 1)/(2x), still 0/0; once more gives eˣ/2 → 1/2. So the limit is 1/2.

Worked Examples

Example 1 — Linear approximation with an over/underestimate justification (NO CALC)

Problem. Let f(x) = ∛x. Use a tangent-line approximation at a convenient point to estimate ∛27.54. Is your estimate an overestimate or an underestimate? Justify.

Strategy. 27 is a perfect cube, so take a = 27.

Solution. f(x) = x^{1/3}, f'(x) = (1/3)x^{−2/3}. - f(27) = 3 - f'(27) = (1/3)(27)^{−2/3} = (1/3)(1/9) = 1/27

L(x) = 3 + (1/27)(x − 27)
L(27.54) = 3 + (1/27)(0.54) = 3 + 0.02 = 3.02

So ∛27.54 ≈ 3.02. (True value ≈ 3.0199.)

Justification. f''(x) = −(2/9)x^{−5/3}, which is negative for x > 0, so f is concave down near x = 27. Because the curve is concave down, the tangent line lies above the curve, so L(27.54) = 3.02 is an overestimate of ∛27.54.

Example 2 — Differentials for propagated error (NO CALC)

Problem. A spherical balloon is measured to have radius r = 10 cm, with a possible measurement error of up to 0.05 cm. Use a differential to estimate the maximum error in the computed volume V = (4/3)πr³.

Strategy. Treat the radius error as dr = 0.05 and propagate it through dV = (dV/dr) dr.

Solution.

dV = 4πr² dr = 4π(10)²(0.05) = 4π·100·0.05 = 20π ≈ 62.8 cm³

So the volume could be off by about 20π ≈ 62.8 cm³. Notice the structure: a 0.05 cm radius error gets amplified by the surface-area factor 4πr².

Example 3 — L'Hôpital, one application (NO CALC)

Problem. Evaluate lim_{x→0} (e^{2x} − 1)/sin x.

Strategy. Check the form first.

Solution. Substituting x = 0: (e⁰ − 1)/sin 0 = 0/0 — indeterminate. ✓ The functions are differentiable near 0 and the denominator's derivative cos x ≠ 0 near 0, so L'Hôpital applies:

lim_{x→0} (e^{2x} − 1)/sin x = lim_{x→0} 2e^{2x}/cos x = 2e⁰/cos 0 = 2/1 = 2

The limit is 2.

Example 4 — Recognize indeterminate vs. not; apply twice when needed (NO CALC)

Problem. Evaluate each limit, applying L'Hôpital only when justified. (i) lim_{x→3} (x² − 9)/(x − 3) (ii) lim_{x→2} (x² − 4)/(x − 1) (iii) lim_{x→0} (1 − cos x)/x²

Solution.

(i) Substitute: (9 − 9)/(3 − 3) = 0/0. Indeterminate ✓ → L'Hôpital:

lim_{x→3} (2x)/1 = 6

(Check: factoring gives (x+3) → 6. Same answer.)

(ii) Substitute: (4 − 4)/(2 − 1) = 0/1 = 0. This is 0 over a nonzero number — NOT indeterminate. The limit is simply 0. Do not apply L'Hôpital here. (If you did, you'd get 2x/1 → 4, which is wrong.)

(iii) Substitute: (1 − 1)/0 = 0/0. Indeterminate ✓ → L'Hôpital:

lim_{x→0} (sin x)/(2x)

Still 0/0, hypotheses still hold → apply again:

lim_{x→0} (cos x)/2 = 1/2

The limit is 1/2.

Common Mistakes

1. Applying L'Hôpital to a non-indeterminate form. Students see a limit of a quotient and reflexively differentiate top and bottom. But if substitution gives 0/5, 5/0, or 3/7, the form is determinate — the answer is 0, the limit is infinite/DNE, or it's just 3/7. Fix: Always substitute first and confirm the form is 0/0 or ±∞/±∞ before invoking the rule.

2. Using the quotient rule instead of separate derivatives. L'Hôpital says lim f/g = lim f'/g' — differentiate numerator and denominator independently. It is not (f/g)' = (f'g − fg')/g². Fix: Write f' over g' as a fresh fraction; never touch the quotient rule here.

3. Getting the over/underestimate direction backwards. "Concave up means the estimate is too big" is a common false memory. Fix: Picture it. Concave up = the curve smiles, bending up away from a tangent line that sits below it → linearization underestimates. Concave down = frown, tangent above → overestimate. Check the sign of f'' at a.

4. Forgetting the (x − a) factor in L(x). Students write L(x) = f(a) + f'(a)·x or plug the target straight into f'. Fix: The linearization is f(a) + f'(a)(x − a); the slope multiplies the displacement from a, not x itself.

5. Stopping L'Hôpital too early — or too late. If the first derivative-ratio is still 0/0 or ∞/∞, apply again. But once the form becomes determinate, stop and substitute. Fix: Re-check the form after every application.

Practice Problems

lim_{x→0} (x − sin x)/x³ = lim_{x→0} (1 − cos x)/(3x²)

= lim_{x→0} (sin x)/(6x)

= lim_{x→0} (cos x)/6 = 1/6

L(x) = 8 + 12(x − 2),  so  L(2.1) = 8 + 12(0.1) = 9.2

11. [NO CALC] A differentiable function has g(4) = 3 and g'(4) = −2. Use a tangent-line approximation to estimate g(4.2).

12. [NO CALC] Evaluate lim_{x→0} (x − sin x)/x³, applying L'Hôpital as many times as needed. Show that each application is justified.

13. [NO CALC — JUSTIFICATION] Let f(x) = x³ and use the tangent line at a = 2 to estimate f(2.1). State the estimate, then determine whether it is an overestimate or underestimate and justify using concavity.

14. [NO CALC — JUSTIFICATION] A student writes: "lim_{x→1} (x² + 3)/(x − 1) is 0/0, so by L'Hôpital it equals lim_{x→1} 2x/1 = 2." Identify the error and find the correct behavior of the limit.

15. [CALC] The volume of a cube is computed from a measured edge s = 12 cm with possible error ds = 0.04 cm. Use a differential to estimate the maximum error in the computed volume, and give the result to one decimal place.

AP Exam Focus

Free Response — Calculator NOT permitted on this question.

The depth of water in a tank, in feet, is modeled by a differentiable function D of time t, measured in hours, for 0 ≤ t ≤ 8. Selected values are given in the table, and it is known that D is concave down on the interval 4 ≤ t ≤ 8.

t (hours)	0	2	4	6	8
D(t) (feet)	3.0	4.6	5.4	5.8	6.0

(a) (2 points) Use the data in the table to find the equation of the tangent line to the graph of D at t = 4, using the average rate of change of D over [4, 6] as an approximation for D'(4).

(b) (2 points) Use your tangent line from part (a) to approximate D(4.5), the depth of the water at time t = 4.5 hours.

(c) (2 points) Is the approximation in part (b) an overestimate or an underestimate of D(4.5)? Give a reason for your answer.

(d) (3 points) A separate model gives the inflow rate R(t) = (e^{2t} − 1)/(sin t) cubic feet per hour for small t > 0. Evaluate lim_{t→0⁺} R(t), showing that the limit is an indeterminate form before applying L'Hôpital's Rule.

Total: 9 points

Model Solution

(a) Approximate the slope by the average rate of change of D over [4, 6]:

D'(4) ≈ [D(6) − D(4)] / (6 − 4) = (5.8 − 5.4)/2 = 0.4/2 = 0.2 feet per hour

The tangent line passes through (4, D(4)) = (4, 5.4) with slope 0.2:

y = 5.4 + 0.2(t − 4)

(b)

D(4.5) ≈ 5.4 + 0.2(4.5 − 4) = 5.4 + 0.2(0.5) = 5.4 + 0.1 = 5.5 feet

(c) The approximation is an overestimate. Because D is concave down on 4 ≤ t ≤ 8, the graph of D lies below its tangent line on that interval. Since t = 4.5 is in this interval, the tangent-line value 5.5 is greater than the actual value D(4.5), so the approximation is an overestimate.

(d) As t → 0⁺, the numerator e^{2t} − 1 → e⁰ − 1 = 0 and the denominator sin t → sin 0 = 0, so R(t) has the indeterminate form 0/0. The numerator and denominator are differentiable near t = 0 and (sin t)' = cos t ≠ 0 for t near 0, so L'Hôpital's Rule applies:

lim_{t→0⁺} (e^{2t} − 1)/(sin t) = lim_{t→0⁺} 2e^{2t}/cos t = 2e⁰/cos 0 = 2/1 = 2

The limit is 2 cubic feet per hour.

Scoring Commentary

• (a) 2 pts: 1 pt for the slope 0.2 (correct average rate of change); 1 pt for a correct tangent-line equation through (4, 5.4). Students lose a point by computing the slope over the wrong interval (e.g., [2,4]) or forgetting the point.
• (b) 2 pts: 1 pt for substituting t = 4.5 into their tangent line; 1 pt for the answer 5.5. Follow-through is allowed from part (a).
• (c) 2 pts: 1 pt for stating "overestimate"; 1 pt for a concavity-based reason ("concave down ⇒ tangent line lies above the curve"). A bare "overestimate" with no reason, or a reason citing increasing/decreasing instead of concavity, earns only the first point.
• (d) 3 pts: 1 pt for showing the form is 0/0 (the indeterminacy must be demonstrated, not assumed); 1 pt for correctly differentiating numerator and denominator separately; 1 pt for the value 2. Applying L'Hôpital without first verifying 0/0 loses the first point — this is the single most common deduction.

lim_{x→0} (x − sin x)/x³ = lim_{x→0} (1 − cos x)/(3x²)

= lim_{x→0} (sin x)/(6x)

= lim_{x→0} (cos x)/6 = 1/6

L(x) = 8 + 12(x − 2),  so  L(2.1) = 8 + 12(0.1) = 9.2