AP Calculus AB · Lesson 3

Tangent-line estimates and L’Hôpital’s rule

Know Know it

Near x=ax = a, a differentiable function stays close to its tangent line. The local linear approximation is L(x)=f(a)+f′(a)(x−a)L(x) = f(a) + f'(a)(x - a), and f(x)≈L(x)f(x) \approx L(x) for xx near aa. The closer xx is to aa, the better the estimate usually is.

Too high or too low? If f′′>0f'' > 0 between aa and xx, the graph is concave up and lies above its tangent line, so L(x)L(x) is an underestimate. If f′′<0f'' < 0 there, the graph is concave down and L(x)L(x) is an overestimate.

L’Hôpital’s rule. Suppose lim⁡x→af(x)=0\lim_{x \to a} f(x) = 0 and lim⁡x→ag(x)=0\lim_{x \to a} g(x) = 0, or both limits are infinite, and ff and gg are differentiable near aa with g′(x)≠0g'(x) \neq 0 near aa (except possibly at aa). Then lim⁡x→af(x)g(x)=lim⁡x→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}, as long as the second limit exists or is infinite. The rule also works for limits as x→∞x \to \infty.

Differentiate the top and the bottom separately. This is not the quotient rule. If the new limit is still 0/0, you may apply the rule again.

Trap Check the form before you use the rule

lim⁡x→0cos⁡xx+1\lim_{x \to 0} \frac{\cos x}{x + 1} substitutes to 11=1\frac{1}{1} = 1. That is the limit. Applying L’Hôpital’s rule anyway gives −sin⁡01=0\frac{-\sin 0}{1} = 0, which is wrong. Use the rule only when substitution gives 00\frac{0}{0} or ∞∞\frac{\infty}{\infty}.

Worked example Easy

A differentiable function has f(3)=7f(3) = 7 and f′(3)=−0.5f'(3) = -0.5. Use the tangent line at x=3x = 3 to estimate f(3.2)f(3.2).

  1. 6.5
  2. 6.9Answer
  3. 7.1
  4. 7.5

How to solve it

  1. Tangent line: L(x)=f(3)+f′(3)(x−3)=7−0.5(x−3)L(x) = f(3) + f'(3)(x - 3) = 7 - 0.5(x - 3).
  2. L(3.2)=7−0.5(0.2)=6.9L(3.2) = 7 - 0.5(0.2) = 6.9.

Why each choice is right or wrong

  • A. Incorrect. 6.5 is 7−0.57 - 0.5: it adds the slope without multiplying it by the step x−3=0.2x - 3 = 0.2.
  • B. Correct. L(3.2)=7+(−0.5)(0.2)=7−0.1=6.9L(3.2) = 7 + (-0.5)(0.2) = 7 - 0.1 = 6.9.
  • C. Incorrect. 7.1 is 7+0.5(0.2)7 + 0.5(0.2), which drops the negative sign of f′(3)f'(3).
  • D. Incorrect. 7.5 is 7+0.57 + 0.5, which both drops the sign and skips multiplying by 0.2.

Explain Explain it

Task verbs: justify, calculate

Justify an over- or underestimate by naming the sign of f′′f'' on the interval, the concavity it gives, and where the tangent line sits: "Because f′′<0f'' < 0 on the interval, the graph is concave down and lies below its tangent line, so the estimate is too high."

Justify using L’Hôpital’s rule by showing that the top and the bottom both approach 0 (or both grow without bound). Write the limits, then write the new quotient of derivatives.

Worked example Medium

What is lim⁡x→0e3x−1sin⁡(2x)\lim_{x \to 0} \frac{e^{3x} - 1}{\sin(2x)}?

  1. 23\frac{2}{3}
  2. 11
  3. 32\frac{3}{2}Answer
  4. 33

How to solve it

  1. Check the form: e0−1=0e^0 - 1 = 0 and sin⁡0=0\sin 0 = 0, so it is 00\frac{0}{0}.
  2. Differentiate top and bottom: 3e3x2cos⁡(2x)\frac{3e^{3x}}{2\cos(2x)}.
  3. Substitute x=0x = 0: 3(1)2(1)=32\frac{3(1)}{2(1)} = \frac{3}{2}.

Why each choice is right or wrong

  • A. Incorrect. 23\frac{2}{3} flips the quotient, putting the derivative of the bottom over the derivative of the top.
  • B. Incorrect. 1 comes from e3xcos⁡(2x)\frac{e^{3x}}{\cos(2x)}, which leaves out the chain-rule factors 3 and 2.
  • C. Correct. Both parts approach 0, so the rule applies: 3e3x2cos⁡(2x)→32\frac{3e^{3x}}{2\cos(2x)} \to \frac{3}{2} as x→0x \to 0.
  • D. Incorrect. 3 keeps the factor 3 from e3xe^{3x} but drops the factor 2 from the chain rule on sin⁡(2x)\sin(2x).

Justify Your turn

Use L’Hôpital’s rule to find lim⁡x→0ln⁡(1+x)x\lim_{x \to 0} \frac{\ln(1 + x)}{x}. Show why the rule can be used.

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Compare with a model answer

A full answer does this

  • Shows the limit has the form 00\frac{0}{0}: the top and the bottom both approach 0 as x→0x \to 0
  • Differentiates the top and the bottom separately to get 1/(1+x)1\frac{1/(1 + x)}{1}
  • Finds that the limit is 1

Model answer

As x→0x \to 0, ln⁡(1+x)→ln⁡1=0\ln(1 + x) \to \ln 1 = 0 and x→0x \to 0, so the limit has the form 00\frac{0}{0} and L’Hôpital’s rule applies. Differentiating the top and the bottom separately gives lim⁡x→01/(1+x)1=11+0=1\lim_{x \to 0} \frac{1/(1 + x)}{1} = \frac{1}{1 + 0} = 1. So the limit is 1.

Apply Apply it

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