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

(a) 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.

(b) 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. ✓

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.

(c) 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.

(d) Common Mistakes

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.

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.

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.

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.

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.

(e) Practice Problems

A differentiable function has g(4) = 3 and g'(4) = −2. Use a tangent-line approximation to estimate g(4.2).

Evaluate lim_{x→0} (x − sin x)/x³, applying L'Hôpital as many times as needed. Show that each application is justified.

[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.

[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.

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.

(f) 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