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Honors Calculus, Unit 1: Limits and continuity

Seven lessons, about 1–2 days each. Get pencil, paper and a graphing calculator (the free Desmos graphing calculator at desmos.com works).

Lesson 1: The tangent problem: from average to instantaneous rate

You will be able to:

The average rate of change of f on [a, b] is (f(b) − f(a)) / (b − a). It is the slope of the secant line through two points on the graph.

To find the rate at one instant, we shrink the interval. The slopes approach a single number: the slope of the tangent line. That "approach" idea is a limit, and it is the whole subject of this unit.

Worked example. f(x) = x². Find the average rate on [1, 3], then estimate the rate at x = 2.

Average: (9 − 1)/(3 − 1) = 4. Near x = 2: on [2, 2.1] the slope is (4.41 − 4)/0.1 = 4.1; on [2, 2.01] it is 4.01; on [2, 2.001] it is 4.001. The slopes approach 4.

Practice set (try each before opening the answers):

  1. Average rate of g(x) = x³ on [1, 2].
  2. A ball's height is s(t) = 16t² feet (falling distance). Average velocity on [1, 1.5]?
  3. Estimate the instantaneous velocity at t = 1 using [1, 1.01].
  4. For f(x) = 1/x, average rate on [1, 2].
  5. Write the slope of the secant from x = 2 to x = 2 + h for f(x) = x², simplified.
Answers and short solutions
  1. (8 − 1)/1 = 7.
  2. 16(2.25 − 1)/0.5 = 40 ft/s.
  3. 16(1.0201 − 1)/0.01 = 32.16, so about 32 ft/s.
  4. (1/2 − 1)/1 = −1/2.
  5. ((2+h)² − 4)/h = (4h + h²)/h = 4 + h. As h shrinks to 0, this approaches 4.

Honors challenge: Show that the secant slope for f(x) = x² between x = a and x = a + h is 2a + h, and explain what the tangent slope must be.

Challenge solution

((a+h)² − a²)/h = (2ah + h²)/h = 2a + h (h ≠ 0). As h → 0 this approaches 2a, so the tangent slope at x = a is 2a.

Lesson 2: Limits from tables and graphs

You will be able to:

We write limx→a f(x) = L when f(x) gets as close to L as we like by taking x close enough to a (but not equal to a). What happens at a doesn't matter.

A limit does not exist (DNE) if the two sides head to different values, if the values grow without bound, or if they oscillate.

Worked example. Find limx→2 (x² − 4)/(x − 2).

At x = 2 the formula gives 0/0, which is undefined. Table: x = 1.9 → 3.9, 1.99 → 3.99, 2.01 → 4.01, 2.1 → 4.1. The limit is 4. (The graph is the line y = x + 2 with a hole at (2, 4).)

Practice set (try each before opening the answers):

  1. Use x = 0.1, 0.01, −0.01 to estimate limx→0 (sin x)/x (radians).
  2. limx→3 (x² − 9)/(x − 3).
  3. limx→0 |x|/x.
  4. limx→0 sin(π/x).
  5. A function has f(5) = 2 but the graph approaches 7 from both sides of x = 5. What are f(5) and limx→5 f(x)?
Answers and short solutions
  1. 0.99833, 0.99998, 0.99998, so the limit is 1.
  2. 6 (it equals x + 3 away from x = 3).
  3. DNE: the value is 1 from the right and −1 from the left.
  4. DNE: the values swing between −1 and 1 forever as x → 0.
  5. f(5) = 2 and the limit is 7. They don't have to match.

Honors challenge: Explain in one paragraph why limx→a f(x) does not depend on f(a).

Challenge solution

The definition only uses x values near a with x ≠ a, so changing or removing the single value f(a) cannot change which number the nearby outputs approach.

Lesson 3: Limit laws and algebra tricks

You will be able to:

If both limits exist, the limit of a sum, difference, product, constant multiple or quotient (with a nonzero denominator) is the sum, difference, product, multiple or quotient of the limits. For polynomials and rational functions with a nonzero denominator, you can just plug in.

If plugging in gives 0/0, the limit may still exist. Simplify first, then plug in.

Worked example. limx→0 (√(x + 9) − 3)/x.

Multiply by the conjugate: (x + 9 − 9)/(x(√(x+9) + 3)) = 1/(√(x+9) + 3) → 1/6.

Practice set (try each before opening the answers):

  1. limx→1 (x² + 3x − 4)/(x − 1).
  2. limx→−2 (x³ + 8)/(x + 2).
  3. limx→4 (1/x − 1/4)/(x − 4).
  4. limx→2 (3x² − x + 5).
  5. limh→0 ((3 + h)² − 9)/h.
Answers and short solutions
  1. Factor: (x + 4)(x − 1)/(x − 1) = x + 4 → 5.
  2. x³ + 8 = (x + 2)(x² − 2x + 4), so the limit is 4 + 4 + 4 = 12.
  3. Combine: (4 − x)/(4x(x − 4)) = −1/(4x) → −1/16.
  4. Plug in: 12 − 2 + 5 = 15.
  5. (6h + h²)/h = 6 + h → 6.

Honors challenge: Given lim f(x) = 3 and lim g(x) = −2 as x → a, find lim (2f(x) − g(x))/(f(x)g(x)) and name each law you use.

Challenge solution

Sum/difference and constant multiple: numerator → 6 + 2 = 8. Product: denominator → −6. Quotient (denominator ≠ 0): 8/(−6) = −4/3.

Lesson 4: One-sided limits, infinite limits and end behavior

You will be able to:

limx→a⁻ uses x values less than a; limx→a⁺ uses values greater than a. The two-sided limit exists only if both one-sided limits exist and are equal.

If f(x) grows without bound as x → a, we write lim = ∞ (the limit still does not exist as a number) and x = a is a vertical asymptote. limx→∞ f(x) = L gives a horizontal asymptote y = L.

Worked example. limx→∞ (3x² + 1)/(2x² − 5).

Divide top and bottom by x²: (3 + 1/x²)/(2 − 5/x²) → 3/2. Horizontal asymptote y = 3/2.

Practice set (try each before opening the answers):

  1. limx→3⁺ 1/(x − 3) and limx→3⁻ 1/(x − 3).
  2. limx→∞ (5x + 2)/(x² + 1).
  3. limx→−∞ (4x³ − x)/(2x³ + 7).
  4. f(x) = x + 1 for x < 2 and f(x) = 5 − x for x ≥ 2. Find both one-sided limits at 2.
  5. limx→0 1/x².
Answers and short solutions
  1. +∞ and −∞. Vertical asymptote x = 3.
  2. 0 (the bottom has the higher degree).
  3. 2.
  4. Left: 3. Right: 3. So the limit is 3.
  5. +∞ from both sides; vertical asymptote x = 0.

Honors challenge: Find limx→∞ (√(x² + 4x) − x).

Challenge solution

Multiply by the conjugate: 4x/(√(x² + 4x) + x) = 4/(√(1 + 4/x) + 1) → 4/2 = 2.

Lesson 5: Continuity and the Intermediate Value Theorem

You will be able to:

f is continuous at a if (1) f(a) is defined, (2) limx→a f(x) exists and (3) they are equal. Breaks are removable (holes), jumps or infinite.

Intermediate Value Theorem: if f is continuous on [a, b] and N is between f(a) and f(b), then some c in (a, b) has f(c) = N.

Worked example. Find k so f is continuous: f(x) = x² for x < 2, and kx + 1 for x ≥ 2.

Need 2² = 2k + 1, so 4 = 2k + 1 and k = 3/2.

Practice set (try each before opening the answers):

  1. Show x³ + x − 1 = 0 has a root between 0 and 1.
  2. Classify the break of (x² − 1)/(x − 1) at x = 1.
  3. Classify the break of the floor function ⌊x⌋ at x = 2.
  4. Where is f(x) = 1/(x² − 4) continuous?
  5. Find a and b so f is continuous: ax + b for x < 1; 3 at x = 1; x² + a for x > 1.
Answers and short solutions
  1. f(0) = −1 < 0 and f(1) = 1 > 0; f is a polynomial (continuous), so by the IVT there is a root in (0, 1).
  2. Removable (a hole at (1, 2)).
  3. Jump discontinuity (left limit 1, right limit 2).
  4. Everywhere except x = 2 and x = −2.
  5. Right side: 1 + a = 3, so a = 2. Left: a + b = 3, so b = 1.

Honors challenge: Use the IVT to prove that at some moment on a 6-hour hike that started at 1,000 ft and ended at 3,000 ft, you were at exactly 2,026 ft. State the assumption you need.

Challenge solution

Altitude is a continuous function of time on [0, 6] with A(0) = 1000 and A(6) = 3000. Since 2026 is between them, the IVT gives a time c with A(c) = 2026. The assumption is continuity: you can't teleport.

Lesson 6: The Squeeze Theorem and trig limits

You will be able to:

Squeeze Theorem: if g(x) ≤ f(x) ≤ h(x) near a and g and h both approach L, then f approaches L too.

Two key facts (x in radians): limx→0 (sin x)/x = 1 and limx→0 (1 − cos x)/x = 0.

Worked example. limx→0 x² sin(1/x).

−1 ≤ sin(1/x) ≤ 1, so −x² ≤ x² sin(1/x) ≤ x². Both outer limits are 0, so the limit is 0.

Practice set (try each before opening the answers):

  1. limx→0 sin(3x)/x.
  2. limx→0 sin(5x)/sin(2x).
  3. limx→0 tan x / x.
  4. limx→∞ (sin x)/x.
  5. limx→0 (1 − cos x)/x².
Answers and short solutions
  1. 3 · sin(3x)/(3x) → 3.
  2. (5x · [sin 5x/5x]) / (2x · [sin 2x/2x]) → 5/2.
  3. (sin x / x)(1/cos x) → 1.
  4. 0, by squeezing between −1/x and 1/x.
  5. Multiply by (1 + cos x): sin²x/(x²(1 + cos x)) → 1 · 1/2 = 1/2.

Honors challenge: Using a unit-circle picture, the areas give sin x ≤ x ≤ tan x for 0 < x < π/2. Derive cos x ≤ (sin x)/x ≤ 1 and conclude the limit.

Challenge solution

Divide sin x ≤ x ≤ tan x by sin x > 0: 1 ≤ x/sin x ≤ 1/cos x. Take reciprocals (flips the inequalities): cos x ≤ sin x/x ≤ 1. As x → 0⁺, cos x → 1, so by the Squeeze Theorem the limit is 1; since sin x / x is even, the left limit is also 1.

Lesson 7: The precise definition of a limit (ε–δ) and Unit 1 test

You will be able to:

limx→a f(x) = L means: for every ε > 0 there is a δ > 0 so that if 0 < |x − a| < δ, then |f(x) − L| < ε. In words: you can force the output within ε of L by keeping the input within δ of a.

Worked example. Prove limx→2 (3x − 1) = 5.

Given ε > 0, choose δ = ε/3. If 0 < |x − 2| < δ, then |(3x − 1) − 5| = 3|x − 2| < 3δ = ε. ∎

Practice set (try each before opening the answers):

  1. Prove limx→1 (5x + 2) = 7. What δ works?
  2. For limx→4 (x/2 + 1) = 3 and ε = 0.1, find the largest δ.
  3. Unit test 1. limx→5 (x² − 25)/(x − 5).
  4. Unit test 2. limx→0 (√(4 + x) − 2)/x.
  5. Unit test 3. limx→∞ (7x² − 3)/(1 − 2x²).
  6. Unit test 4. limx→0 sin(4x)/(3x).
  7. Unit test 5. Find c so f(x) = cx² for x ≤ 3 and 2x + c for x > 3 is continuous.
  8. Unit test 6. Show cos x = x has a solution in (0, 1).
Answers and short solutions
  1. δ = ε/5, since |5x + 2 − 7| = 5|x − 1|.
  2. |x/2 − 2| = |x − 4|/2 < 0.1 gives δ = 0.2.
  3. 10.
  4. 1/4.
  5. −7/2.
  6. 4/3.
  7. 9c = 6 + c, so c = 3/4.
  8. g(x) = cos x − x is continuous, g(0) = 1 > 0, g(1) = cos 1 − 1 ≈ −0.46 < 0, so the IVT gives a root.

Honors challenge: Prove limx→3 x² = 9. (Hint: first require δ ≤ 1 so |x + 3| < 7.)

Challenge solution

|x² − 9| = |x − 3||x + 3|. If δ ≤ 1 then 2 < x < 4, so |x + 3| < 7. Choose δ = min(1, ε/7). Then |x² − 9| < 7 · ε/7 = ε. ∎

AP® Calculus AB exam prep (optional)

On the AP® Calculus AB exam, limits appear in multiple-choice questions and as justifications in free-response questions (for example, using the definition of continuity or the IVT). Practice writing a full sentence: "Because f is continuous on [a, b] and f(a) < N < f(b), the Intermediate Value Theorem guarantees …".

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Plans: $35/month per student or $69/month per family (see plans). Optional live 1:1 tutoring: $25 for your first hour, then $35/hour, or 4 hours for $130 (tutoring).

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