Tasks 05-01 - Limits & Continuity Through Graphs

Section 05: Differential Calculus

A: Skills

Problem 1: Limits from a Graph (x)

The figure shows the graph of a function \(f\). An open circle marks a point that does not belong to the graph, a filled circle a point that does. The dashed line is a vertical asymptote.

  1. Determine \(\lim_{x \to -1^-} f(x)\), \(\lim_{x \to -1^+} f(x)\) and \(f(-1)\). Decide whether \(\lim_{x \to -1} f(x)\) exists.

  2. Determine \(\lim_{x \to 1} f(x)\) and \(f(1)\). Decide whether \(f\) is continuous at \(x = 1\).

  3. Explain why \(\lim_{x \to 2} f(x) = 3\) and decide whether \(f\) is continuous at \(x = 2\).

  4. Determine both one-sided limits at \(x = 4\). Decide whether \(f(4)\) is defined.

  5. List all \(x\) in \([-4; 8]\) at which \(f\) is not continuous and name the type of discontinuity (removable, jump or infinite) at each of them.

Trace the graph from each side towards the \(x\)-value. The \(y\)-value the curve approaches is the one-sided limit. The filled circle gives the function value.

  1. At \(x = -1\):

    • From the left, the line approaches the open circle at \((-1, 3)\): \(\lim_{x \to -1^-} f(x) = 3\)
    • From the right, the parabola starts at the filled circle \((-1, 0)\): \(\lim_{x \to -1^+} f(x) = 0\)
    • \(f(-1) = 0\) (filled circle)
    • \(3 \neq 0\): the one-sided limits differ, so \(\lim_{x \to -1} f(x)\) does not exist.

  2. At \(x = 1\):

    • From both sides, the parabola approaches the open circle at \((1, 0)\): \(\lim_{x \to 1} f(x) = 0\)
    • \(f(1) = 2\) (the filled circle above the gap)
    • The limit exists, but \(0 \neq 2\). Condition 3 (limit = function value) fails, so \(f\) is not continuous at \(x = 1\).

  3. At \(x = 2\), the parabola from the left and the falling line from the right both approach the filled circle \((2, 3)\):

    \[\lim_{x \to 2^-} f(x) = 3 = \lim_{x \to 2^+} f(x)\]

    Both one-sided limits are equal, so \(\lim_{x \to 2} f(x) = 3\). Also \(f(2) = 3\), so all three conditions hold: \(f\) is continuous at \(x = 2\). The graph has a corner there, but no gap and no jump.

  4. At \(x = 4\):

    • From the left, the line approaches the open circle \((4, 1)\): \(\lim_{x \to 4^-} f(x) = 1\)
    • From the right, the curve rises along the asymptote without bound: \(\lim_{x \to 4^+} f(x) = +\infty\)
    • There is no filled circle at \(x = 4\), so \(f(4)\) is not defined.

  5. Discontinuities in \([-4; 8]\):

\(x\) What happens Type
\(-1\) left limit 3, right limit 0 jump discontinuity
\(1\) limit 0, but \(f(1) = 2\) removable discontinuity
\(4\) right limit \(+\infty\), \(f(4)\) not defined infinite discontinuity

At every other \(x\) in \([-4; 8]\) the graph can be drawn without lifting the pen, so \(f\) is continuous there (including the corner at \(x = 2\)).

Problem 2: One-Sided Limits of a Discount Policy (x)

A wholesaler of office supplies grants a discount on the list price that depends on the number \(x\) of packs in one order. \(D(x)\) is the discount in percent:

\[D(x) = \begin{cases} 0 & \text{if } 0 < x < 30 \\ 4 & \text{if } 30 \leq x < 75 \\ 6 & \text{if } x = 75 \\ 8 & \text{if } 75 < x < 150 \\ 12 & \text{if } x \geq 150 \end{cases}\]

  1. Determine \(\lim_{x \to 30^-} D(x)\), \(\lim_{x \to 30^+} D(x)\) and \(D(30)\).

  2. Determine \(\lim_{x \to 75^-} D(x)\), \(\lim_{x \to 75^+} D(x)\) and \(D(75)\).

  3. Determine \(\lim_{x \to 150^-} D(x)\) and \(\lim_{x \to 150^+} D(x)\).

  4. Check the three conditions for continuity at \(x = 30\), \(x = 75\) and \(x = 150\) and name the type of each discontinuity.

  5. Sketch the graph of \(D\) for \(0 < x < 200\). Mark open and filled circles.

  6. The list price is 10 EUR per pack. Compute the invoice amount for 74, 75 and 76 packs. Explain with the one-sided limits of the invoice amount at \(x = 75\) what a customer who needs 74 packs should do.

  1. Left of 30, the discount is 0 %, from 30 on it is 4 %:

    • \(\lim_{x \to 30^-} D(x) = 0\)
    • \(\lim_{x \to 30^+} D(x) = 4\)
    • \(D(30) = 4\) (the second line includes \(x = 30\))

  2. Left of 75, the discount is 4 %, right of 75 it is 8 %:

    • \(\lim_{x \to 75^-} D(x) = 4\)
    • \(\lim_{x \to 75^+} D(x) = 8\)
    • \(D(75) = 6\) (the special line for exactly 75 packs)

  3. \(\lim_{x \to 150^-} D(x) = 8\) and \(\lim_{x \to 150^+} D(x) = 12\).

  4. Continuity check:

Point 1. value defined 2. limit exists 3. limit = value Result
\(x = 30\) \(D(30) = 4\) no (\(0 \neq 4\)) no jump discontinuity
\(x = 75\) \(D(75) = 6\) no (\(4 \neq 8\)) no jump discontinuity
\(x = 150\) \(D(150) = 12\) no (\(8 \neq 12\)) no jump discontinuity

\(D\) is not continuous at any of the three thresholds. Between the thresholds, \(D\) is constant and therefore continuous.

  1. The graph is a step function:

    • a horizontal segment at height 0 from \(x = 0\) to \(x = 30\), with an open circle at \((30, 0)\)
    • a segment at height 4 from a filled circle at \((30, 4)\) to an open circle at \((75, 4)\)
    • a single filled point at \((75, 6)\)
    • a segment at height 8 from an open circle at \((75, 8)\) to an open circle at \((150, 8)\)
    • a segment at height 12 from a filled circle at \((150, 12)\) to \(x = 200\)

  2. Invoice amount: \(I(x) = 10x \cdot \left(1 - \frac{D(x)}{100}\right)\) EUR.

    • 74 packs: \(740 \cdot 0.96 = 710.40\) EUR
    • 75 packs: \(750 \cdot 0.94 = 705.00\) EUR
    • 76 packs: \(760 \cdot 0.92 = 699.20\) EUR

    One-sided limits of the invoice amount at \(x = 75\):

    \[\lim_{x \to 75^-} I(x) = 750 \cdot 0.96 = 720 \text{ EUR}, \qquad \lim_{x \to 75^+} I(x) = 750 \cdot 0.92 = 690 \text{ EUR}\]

    The invoice amount jumps down by 30 EUR at 75 packs. A customer who needs 74 packs should order 76 packs: two packs more cost 11.20 EUR less. Discount steps on the whole order make “more for less” possible.

Problem 3: Limits by Factoring (x)

Compute the following limits. Factor and cancel first where direct substitution gives \(\frac{0}{0}\).

  1. \(\lim_{x \to 4} \frac{x^2 - 16}{x^2 - 3x - 4}\)

  2. \(\lim_{x \to -5} \frac{x^2 + 3x - 10}{x + 5}\)

  3. \(\lim_{x \to 9} \frac{x - 9}{\sqrt{x} - 3}\)

  4. \(\lim_{x \to 0} \frac{x^3 + 4x^2}{2x^2}\)

  5. Consider the function \(g(x) = \frac{x^3 - 1}{x^2 + 2x - 3}\).

    (1) Simplify \(g\) by factoring numerator and denominator (difference of cubes) and give the domain of \(g\).

    (2) Determine \(\lim_{x \to 1} g(x)\) and both one-sided limits at \(x = -3\). Name the type of discontinuity at \(x = 1\) and at \(x = -3\).

    (3) Give the continuous extension of \(g\) at \(x = 1\) (the function that agrees with \(g\) everywhere else and is continuous at \(x = 1\)). Explain why \(g\) cannot be made continuous at \(x = -3\).

First try direct substitution. If it gives \(\frac{0}{0}\), a common factor is hidden in numerator and denominator.

  1. Substitution gives \(\frac{0}{0}\). Factor:

    \[\frac{x^2 - 16}{x^2 - 3x - 4} = \frac{(x - 4)(x + 4)}{(x - 4)(x + 1)} = \frac{x + 4}{x + 1} \quad (x \neq 4)\]

    \[\lim_{x \to 4} \frac{x + 4}{x + 1} = \frac{8}{5} = 1.6\]

  2. Substitution gives \(\frac{25 - 15 - 10}{0} = \frac{0}{0}\). Factor: \(x^2 + 3x - 10 = (x + 5)(x - 2)\).

    \[\lim_{x \to -5} \frac{(x + 5)(x - 2)}{x + 5} = \lim_{x \to -5} (x - 2) = -7\]

  3. Substitution gives \(\frac{0}{0}\). Multiply by the conjugate \(\sqrt{x} + 3\) (difference of squares, \((\sqrt{x} - 3)(\sqrt{x} + 3) = x - 9\)):

    \[\frac{x - 9}{\sqrt{x} - 3} \cdot \frac{\sqrt{x} + 3}{\sqrt{x} + 3} = \frac{(x - 9)(\sqrt{x} + 3)}{x - 9} = \sqrt{x} + 3 \quad (x \neq 9)\]

    \[\lim_{x \to 9} (\sqrt{x} + 3) = 3 + 3 = 6\]

  4. Substitution gives \(\frac{0}{0}\). Factor out \(x^2\):

    \[\frac{x^3 + 4x^2}{2x^2} = \frac{x^2(x + 4)}{2x^2} = \frac{x + 4}{2} \quad (x \neq 0), \qquad \lim_{x \to 0} \frac{x + 4}{2} = 2\]

    1. Numerator (difference of cubes) and denominator:

    \[x^3 - 1 = (x - 1)(x^2 + x + 1), \qquad x^2 + 2x - 3 = (x - 1)(x + 3)\]

    \[g(x) = \frac{(x - 1)(x^2 + x + 1)}{(x - 1)(x + 3)} = \frac{x^2 + x + 1}{x + 3} \quad (x \neq 1)\]

    Domain: the original denominator is zero at \(x = 1\) and \(x = -3\), so \(D = \mathbb{R} \setminus \{-3, 1\}\).

    1. At \(x = 1\), use the simplified form:

    \[\lim_{x \to 1} g(x) = \frac{1 + 1 + 1}{1 + 3} = \frac{3}{4}\]

    The limit exists, but \(g(1)\) is not defined: removable discontinuity (hole at \(\left(1, \frac{3}{4}\right)\)).

    At \(x = -3\), the numerator of the simplified form is \(9 - 3 + 1 = 7 > 0\) and the denominator \(x + 3\) goes to 0:

    • from the left (\(x < -3\)), \(x + 3\) is a small negative number: \(\lim_{x \to -3^-} g(x) = -\infty\)
    • from the right (\(x > -3\)), \(x + 3\) is a small positive number: \(\lim_{x \to -3^+} g(x) = +\infty\)

    Infinite discontinuity (vertical asymptote \(x = -3\)).

    1. Continuous extension at \(x = 1\):

    \[\tilde{g}(x) = \begin{cases} \frac{x^3 - 1}{x^2 + 2x - 3} & \text{if } x \neq 1, \; x \neq -3 \\ \frac{3}{4} & \text{if } x = 1 \end{cases}\]

    At \(x = -3\), the one-sided limits are \(-\infty\) and \(+\infty\). No number can be the limit, so condition 2 fails whatever value we choose for \(g(-3)\).

Problem 4: Limits at Infinity (x)

Compute the following limits. For every finite limit, give the horizontal asymptote of the graph.

  1. \(\lim_{x \to \infty} \frac{5x^2 + 2}{2x^2 - 7x}\)

  2. \(\lim_{x \to \infty} \frac{3x + 7}{x^2 + 1}\)

  3. \(\lim_{x \to -\infty} \frac{x^3 - 2x}{4x^2 + 1}\)

  4. \(\lim_{x \to \infty} \frac{6x - 1}{3 - 2x}\)

  5. \(\lim_{x \to \infty} \frac{9x + 600}{x}\)

  6. \(\lim_{t \to \infty} \left(40 - 25e^{-0.3t}\right)\)

For rational functions, compare the degrees of numerator and denominator: equal degrees give the ratio of the leading coefficients, a smaller numerator degree gives 0, a larger numerator degree gives \(\pm\infty\).

  1. Both degrees are 2: \(\lim_{x \to \infty} \frac{5x^2 + 2}{2x^2 - 7x} = \frac{5}{2} = 2.5\). Horizontal asymptote \(y = 2.5\).

  2. Degree 1 < degree 2: the limit is \(0\). Horizontal asymptote \(y = 0\).

  3. Degree 3 > degree 2: the function grows without bound in absolute value. For large \(|x|\), the fraction behaves like \(\frac{x^3}{4x^2} = \frac{x}{4}\), which goes to \(-\infty\) for \(x \to -\infty\). So the limit is \(-\infty\) (no horizontal asymptote).

  4. Both degrees are 1: ratio of the leading coefficients \(\frac{6}{-2} = -3\). The limit is \(-3\), horizontal asymptote \(y = -3\).

    Typical mistake: taking \(\frac{6}{3} = 2\). The leading coefficient of \(3 - 2x\) is \(-2\), the coefficient of \(x\), not the first number written.

  5. Split the fraction: \(\frac{9x + 600}{x} = 9 + \frac{600}{x}\). Since \(\frac{600}{x} \to 0\), the limit is \(9\). Horizontal asymptote \(y = 9\). (This is the pattern of an average cost with fixed costs 600 and variable costs 9 per unit.)

  6. For \(t \to \infty\), \(e^{-0.3t} = \frac{1}{e^{0.3t}} \to 0\), so the limit is \(40 - 25 \cdot 0 = 40\). Horizontal asymptote \(y = 40\).

Problem 5: Continuity with the Three Conditions (xx)

Check the three conditions for continuity at the given point. If a function is not continuous there, name the type of discontinuity and decide whether changing one function value makes it continuous. If so, give that value.

  1. \(f(x) = \begin{cases} 2x + 1 & \text{if } x < 2 \\ 7 - x & \text{if } x \geq 2 \end{cases}\) at \(x = 2\)

  2. \(g(x) = \begin{cases} x^2 - 4 & \text{if } x \leq 3 \\ 2x & \text{if } x > 3 \end{cases}\) at \(x = 3\)

  3. \(h(x) = \begin{cases} \frac{x^2 - 5x + 6}{x - 3} & \text{if } x \neq 3 \\ 2 & \text{if } x = 3 \end{cases}\) at \(x = 3\)

  4. \(k(x) = \begin{cases} x + 2 & \text{if } x \leq 1 \\ \frac{4}{x - 1} & \text{if } x > 1 \end{cases}\) at \(x = 1\)

The three conditions at \(x = a\): (1) \(f(a)\) is defined, (2) \(\lim_{x \to a} f(x)\) exists (both one-sided limits are equal), (3) \(\lim_{x \to a} f(x) = f(a)\).

  1. At \(x = 2\):

    1. \(f(2) = 7 - 2 = 5\) ✓
    2. \(\lim_{x \to 2^-} f(x) = 2 \cdot 2 + 1 = 5\) and \(\lim_{x \to 2^+} f(x) = 7 - 2 = 5\), so \(\lim_{x \to 2} f(x) = 5\) ✓
    3. \(5 = 5\) ✓

    \(f\) is continuous at \(x = 2\).

  2. At \(x = 3\):

    1. \(g(3) = 9 - 4 = 5\) ✓
    2. \(\lim_{x \to 3^-} g(x) = 5\) and \(\lim_{x \to 3^+} g(x) = 2 \cdot 3 = 6\). \(5 \neq 6\), so the limit does not exist ✗

    Jump discontinuity (jump of 1). Changing \(g(3)\) does not help: whatever value we choose, the two sides still approach different values.

  3. At \(x = 3\):

    1. \(h(3) = 2\) ✓
    2. Substitution into the fraction gives \(\frac{0}{0}\). Factor: \(\frac{(x - 2)(x - 3)}{x - 3} = x - 2\) for \(x \neq 3\), so \(\lim_{x \to 3} h(x) = 1\) ✓
    3. \(1 \neq 2\) ✗

    Removable discontinuity. Changing the function value to \(h(3) = 1\) makes \(h\) continuous at \(x = 3\).

  4. At \(x = 1\):

    1. \(k(1) = 1 + 2 = 3\) ✓
    2. \(\lim_{x \to 1^-} k(x) = 3\), but for \(x \to 1^+\) the denominator \(x - 1\) is a small positive number: \(\lim_{x \to 1^+} k(x) = +\infty\) ✗

    Infinite discontinuity (vertical asymptote \(x = 1\) on the right side). No function value can repair it.

B: Set Up & Check

Problem 6: From Text to Limits (xx)

Translate each statement into mathematics. Then answer the check question: it tells you whether your translation works.

  1. “A company makes ice cubes for bars and events. Up to 2,000 bags per week, its costs are \(C(x) = 700 + 0.8x\) EUR for \(x\) bags. For more than 2,000 bags it rents a freezer truck, and then \(C(x) = 1050 + 0.8x\) EUR.” The manager says: “At 2,000 bags, our weekly costs jump by 350 EUR.” Write this statement with one-sided limits.

    Check: Compute both one-sided limits at \(x = 2000\) and \(C(2000)\). Is the difference of the one-sided limits 350 EUR?

  2. “A mobile network operator sells data volume to companies. Up to 50 GB per month, the price is 1.20 EUR per GB, so \(B(x) = 1.2x\) EUR. For more than 50 GB, the price is \(B(x) = a + 0.8x\) EUR.” The operator says: “Our tariff is continuous at 50 GB.” Write this statement as the three conditions for continuity and determine \(a\).

    Check: Compute \(B(49.9)\) and \(B(50.1)\) with your value of \(a\). Are they close to each other?

  3. “A publisher prints a new cookbook. Printing plates and design cost 18,000 EUR, and each copy costs another 4.50 EUR to print.” The sales manager says: “In the long run, our average cost approaches 4.50 EUR per copy.” Write this statement as a limit.

    Check: Compute the average cost for 1,000, 10,000 and 100,000 copies. Is the average cost ever exactly 4.50 EUR?

  4. “A nut roastery’s daily revenue is \(R(x) = 12x - 0.02x^2\) EUR for \(x\) bags of roasted almonds. Its spreadsheet computes the price per bag as \(\frac{R(x)}{x}\) and shows an error for \(x = 0\).” The analyst says: “The formula is not defined at \(x = 0\), but its values approach 12 EUR.” Write this statement as a limit and name the type of discontinuity at \(x = 0\).

    Check: Compute \(\frac{R(x)}{x}\) for \(x = 1\), \(x = 0.1\) and \(x = 0.01\). Do the values approach 12?

  1. The jump is the difference between the value approached from the right and the value approached from the left:

    \[\lim_{x \to 2000^+} C(x) - \lim_{x \to 2000^-} C(x) = 350\]

    Check: yes. \(\lim_{x \to 2000^-} C(x) = 700 + 1600 = 2{,}300\) EUR, \(\lim_{x \to 2000^+} C(x) = 1050 + 1600 = 2{,}650\) EUR, and \(2650 - 2300 = 350\) ✓. \(C(2000) = 2{,}300\) EUR (the first formula includes \(x = 2000\)). The two-sided limit at \(x = 2000\) does not exist: jump discontinuity.

    Typical mistake: computing the jump as \(C(2001) - C(2000) = 2650.80 - 2300 = 350.80\) EUR. This difference also contains the variable costs of one more bag. The jump itself is the difference of the one-sided limits.

  2. The three conditions at \(x = 50\):

    1. \(B(50) = 1.2 \cdot 50 = 60\) is defined
    2. \(\lim_{x \to 50^-} B(x) = \lim_{x \to 50^+} B(x)\), i.e. \(60 = a + 0.8 \cdot 50 = a + 40\)
    3. \(\lim_{x \to 50} B(x) = B(50) = 60\)

    From condition 2: \(a + 40 = 60\), so \(a = 20\) EUR.

    Check: yes. \(B(49.9) = 1.2 \cdot 49.9 = 59.88\) EUR and \(B(50.1) = 20 + 0.8 \cdot 50.1 = 60.08\) EUR. Both are close to 60 EUR, the tariff has no jump.

    Typical mistake: setting the prices per GB equal, \(1.2 = 0.8\), and concluding that the tariff can never be continuous. Continuity compares the amounts \(B(x)\) near 50 GB, not the prices per GB.

  3. With \(x\) copies: \(C(x) = 18000 + 4.5x\) and \(AC(x) = \frac{C(x)}{x} = \frac{18000}{x} + 4.5\) EUR per copy. The statement:

    \[\lim_{x \to \infty} AC(x) = 4.5\]

    Check: \(AC(1000) = 18 + 4.5 = 22.50\) EUR, \(AC(10000) = 1.8 + 4.5 = 6.30\) EUR, \(AC(100000) = 0.18 + 4.5 = 4.68\) EUR. The values approach 4.50 EUR, but never reach it: \(\frac{18000}{x} > 0\) for every \(x > 0\).

    Typical mistake: writing \(\lim_{x \to \infty} C(x) = 4.5\). The total costs grow without bound. Only the costs per copy approach 4.50 EUR.

  4. \(\frac{R(x)}{x} = \frac{x(12 - 0.02x)}{x} = 12 - 0.02x\) for \(x \neq 0\). The statement:

    \[\lim_{x \to 0^+} \frac{R(x)}{x} = 12\]

    The limit exists, but the formula has no value at \(x = 0\): removable discontinuity (a hole at \((0, 12)\)). 12 EUR is the maximum price: at this price, the demand would drop to 0 bags.

    Check: yes. \(\frac{R(1)}{1} = 11.98\), \(\frac{R(0.1)}{0.1} = 11.998\), \(\frac{R(0.01)}{0.01} = 11.9998\) EUR.

    Typical mistake: “The formula is not defined at \(x = 0\), so the graph has a vertical asymptote there.” A vertical asymptote needs values that grow without bound. Here the factor \(x\) cancels, and the values approach a finite number.

Problem 7: One Paragraph, All Conditions (xx)

A rental company rents out moving vans for one day. For a trip of \(x\) km, it charges 0.60 EUR per km up to 150 km. For more than 150 km, it charges a fixed amount \(a\) plus 0.25 EUR for every kilometre of the trip, and \(a\) is chosen so that the price does not jump at 150 km. For trips of more than 600 km, the company adds a one-time fee of 40 EUR for an extra technical check. The manager wants to know what the price per km approaches for very long trips.

List every condition and every function in this paragraph as a mathematical statement. Do not solve anything.

Check: An assistant writes the condition for \(a\) as \(0.60 = 0.25\). Name the mistake.

  1. Price for short trips: \(P(x) = 0.6x\) EUR for \(0 \leq x \leq 150\)
  2. Price for medium trips: \(P(x) = a + 0.25x\) EUR for \(150 < x \leq 600\)
  3. Price for long trips: \(P(x) = a + 40 + 0.25x\) EUR for \(x > 600\)
  4. No jump at 150 km (three conditions): \(P(150) = 0.6 \cdot 150\) is defined, \(\lim_{x \to 150^-} P(x) = \lim_{x \to 150^+} P(x)\), i.e. \(0.6 \cdot 150 = a + 0.25 \cdot 150\), and this limit equals \(P(150)\)
  5. Jump of 40 EUR at 600 km: \(\lim_{x \to 600^+} P(x) - \lim_{x \to 600^-} P(x) = 40\)
  6. Price per km: \(A(x) = \frac{P(x)}{x}\) for \(x > 0\)
  7. Very long trips: \(\lim_{x \to \infty} A(x)\)

Check: the assistant compares the prices per km. Continuity at 150 km is a condition on the total price \(P(x)\): the price for a trip just under 150 km and the price for a trip just over 150 km must approach the same amount, \(0.6 \cdot 150 = a + 0.25 \cdot 150\).

Typical mistake: writing the jump at 600 km as \(P(601) - P(600) = 40\). This difference also contains the price of one more kilometre. The jump is the difference of the one-sided limits.

C: Applied

Problem 8: A New Income Tax (xx)

A country introduces a new income tax with the following marginal tax rates (the marginal tax rate is the rate on the next euro of income):

Part of the yearly income Marginal tax rate
up to 12,000 EUR 0 %
above 12,000 EUR up to 40,000 EUR 15 %
above 40,000 EUR up to 90,000 EUR 30 %
above 90,000 EUR 45 %

Each rate applies only to the part of the income inside its bracket. \(T(x)\) is the tax in EUR for a yearly income of \(x\) EUR. Round percentages to one decimal place.

  1. Give \(T(x)\) as a piecewise function with four pieces.

(for verification: \(T(40{,}000) = 4{,}200\) EUR and \(T(90{,}000) = 19{,}200\) EUR)

  1. Investigate with the three conditions whether \(T\) is continuous at 12,000 EUR, 40,000 EUR and 90,000 EUR.

  2. Determine the one-sided limits of the marginal tax rate \(r(x)\) at \(x = 40{,}000\) and decide whether \(r\) is continuous there. An employee earns 39,900 EUR and gets a raise to 40,100 EUR. They fear: “From now on, 30 % of my income goes to taxes.” Compute the extra tax on their raise and explain with b) why their fear is wrong.

  3. The effective tax rate is \(E(x) = \frac{T(x)}{x}\) (the share of the whole income that goes to taxes). Compute \(E(40{,}000)\) and \(E(90{,}000)\) in percent. Show that \(E(x) = 0.45 - \frac{21300}{x}\) for \(x > 90{,}000\). Compute \(\lim_{x \to \infty} E(x)\) and interpret the result.

  4. A party proposes a simpler rule: incomes above 40,000 EUR (up to 90,000 EUR) pay 30 % on the whole income above 12,000 EUR. Incomes up to 40,000 EUR are taxed as before. Determine the one-sided limits of the new tax at \(x = 40{,}000\). Determine the income above 40,000 EUR at which the net income (income minus tax) is as high as at exactly 40,000 EUR, and explain what the proposal means for employees who earn a little more than 40,000 EUR.

(for verification: 46,000 EUR)

a) Tax in a bracket = tax at the lower end of the bracket + rate · income above the lower end:

  • \(T(12{,}000) = 0\)
  • \(T(40{,}000) = 0.15 \cdot (40{,}000 - 12{,}000) = 0.15 \cdot 28{,}000 = 4{,}200\) EUR
  • \(T(90{,}000) = 4{,}200 + 0.30 \cdot 50{,}000 = 4{,}200 + 15{,}000 = 19{,}200\) EUR

\[T(x) = \begin{cases} 0 & \text{if } 0 \leq x \leq 12000 \\ 0.15(x - 12000) & \text{if } 12000 < x \leq 40000 \\ 4200 + 0.3(x - 40000) & \text{if } 40000 < x \leq 90000 \\ 19200 + 0.45(x - 90000) & \text{if } x > 90000 \end{cases}\]

b) At each bracket boundary, the value belongs to the lower piece.

Boundary \(T\) at the boundary Left limit Right limit
12,000 \(0\) \(0\) \(0.15 \cdot 0 = 0\)
40,000 \(4{,}200\) \(0.15 \cdot 28{,}000 = 4{,}200\) \(4{,}200 + 0.3 \cdot 0 = 4{,}200\)
90,000 \(19{,}200\) \(4{,}200 + 0.3 \cdot 50{,}000 = 19{,}200\) \(19{,}200 + 0.45 \cdot 0 = 19{,}200\)

At every boundary, (1) the value is defined, (2) both one-sided limits are equal, so the limit exists, and (3) the limit equals the value. \(T\) is continuous at all three boundaries. This is no accident: each piece starts at the tax reached at the end of the previous piece.

c) The marginal tax rate is a step function: \(r(x) = 15\) % just below 40,000 EUR and \(r(x) = 30\) % just above:

\[\lim_{x \to 40000^-} r(x) = 15\,\%, \qquad \lim_{x \to 40000^+} r(x) = 30\,\%\]

The one-sided limits differ, so \(r\) has a jump discontinuity at 40,000 EUR.

Extra tax on the raise:

  • \(T(39{,}900) = 0.15 \cdot 27{,}900 = 4{,}185\) EUR
  • \(T(40{,}100) = 4{,}200 + 0.3 \cdot 100 = 4{,}230\) EUR
  • Extra tax: \(4{,}230 - 4{,}185 = 45\) EUR on a raise of 200 EUR

Their fear is wrong. The rate jumps, but the tax \(T\) is continuous (b)): only the 100 EUR above 40,000 EUR are taxed at 30 %, the rest of their income is taxed as before. Their net income still rises by \(200 - 45 = 155\) EUR.

d) Effective tax rates:

  • \(E(40{,}000) = \frac{4{,}200}{40{,}000} = 0.105 = 10.5\) %
  • \(E(90{,}000) = \frac{19{,}200}{90{,}000} \approx 0.2133 = 21.3\) %

The effective rate is much lower than the marginal rate, because the lower parts of the income are taxed at lower rates.

For \(x > 90{,}000\): \(T(x) = 19{,}200 + 0.45x - 40{,}500 = 0.45x - 21{,}300\), so

\[E(x) = \frac{0.45x - 21300}{x} = 0.45 - \frac{21300}{x} \quad \checkmark\]

Since \(\frac{21300}{x} \to 0\) for \(x \to \infty\):

\[\lim_{x \to \infty} E(x) = 0.45 = 45\,\%\]

Interpretation: for very high incomes, the effective tax rate comes closer and closer to the top rate of 45 %. It never reaches it, because \(\frac{21300}{x} > 0\): the first 90,000 EUR of every income are taxed at lower rates.

e) New tax for \(40{,}000 < x \leq 90{,}000\): \(T_2(x) = 0.3(x - 12000)\). Below 40,000 EUR nothing changes.

\[\lim_{x \to 40000^-} T_2(x) = 0.15 \cdot 28{,}000 = 4{,}200 \text{ EUR}, \qquad \lim_{x \to 40000^+} T_2(x) = 0.3 \cdot 28{,}000 = 8{,}400 \text{ EUR}\]

The tax jumps by 4,200 EUR at 40,000 EUR (jump discontinuity).

Net income at exactly 40,000 EUR: \(40{,}000 - 4{,}200 = 35{,}800\) EUR. Above 40,000 EUR the net income is \(x - 0.3(x - 12000) = 0.7x + 3600\):

\[0.7x + 3600 = 35800 \iff 0.7x = 32200 \iff x = 46{,}000 \text{ EUR}\]

For every income between 40,000 EUR and 46,000 EUR, the net income is lower than at exactly 40,000 EUR. An employee with 40,000 EUR who gets a raise of up to 6,000 EUR has less money than before. Such a tax punishes a raise and pushes people to keep their income just below the threshold. This is why tax functions are built to be continuous.

Problem 9: Three Printing Methods (xx)

A printing company prints customer magazines (magazines that companies send to their customers). For an order of \(x\) copies, it can choose between three methods:

  • Digital printing: 80 EUR for preparing the files, then 1.40 EUR per copy
  • Sheet-fed offset printing (printing plates are needed): 600 EUR for the plates, then 0.60 EUR per copy
  • Web offset printing (printing on large paper rolls): 2,400 EUR for setting up the machine, then 0.24 EUR per copy

  1. Give the cost functions \(C_D(x)\), \(C_S(x)\) and \(C_W(x)\) of the three methods. Compute the costs of all three methods for orders of 500, 2,000 and 8,000 copies and decide which method is the cheapest in each case.

  2. Determine the order sizes at which digital printing and sheet-fed offset printing cost the same, and at which sheet-fed offset printing and web offset printing cost the same. The company always uses the cheapest method. Give its cost function \(C(x)\) as a piecewise function.

(for verification: 650 copies and 5,000 copies)

  1. Show with the three conditions that \(C\) is continuous at both switching points. Explain what this means for a customer whose order grows slowly.

  2. The production manager uses a simpler rule: digital printing below 650 copies, sheet-fed offset printing from 650 to 3,999 copies and web offset printing from 4,000 copies on. Let \(C_R(x)\) be the costs under this rule. Determine the one-sided limits of \(C_R\) at \(x = 4000\) and explain the jump. Compute the extra costs that the rule causes for an order of 4,500 copies.

  3. The company charges 1.00 EUR per copy. Explain why it makes a loss on every order it prints digitally. Determine the smallest order that brings a profit when the company uses the cheapest method.

a) Cost functions in EUR:

\[C_D(x) = 80 + 1.4x, \qquad C_S(x) = 600 + 0.6x, \qquad C_W(x) = 2400 + 0.24x\]

Copies Digital Sheet-fed offset Web offset Cheapest
500 780 900 2,520 digital
2,000 2,880 1,800 2,880 sheet-fed offset
8,000 11,280 5,400 4,320 web offset

b) Equal costs:

\[80 + 1.4x = 600 + 0.6x \iff 0.8x = 520 \iff x = 650\]

\[600 + 0.6x = 2400 + 0.24x \iff 0.36x = 1800 \iff x = 5000\]

(Digital and web offset cost the same at \(1.16x = 2320\), i.e. \(x = 2000\), but there sheet-fed offset is cheaper, see a).)

The method with the smaller costs per copy wins for larger orders:

\[C(x) = \begin{cases} 80 + 1.4x & \text{if } 0 < x \leq 650 \\ 600 + 0.6x & \text{if } 650 < x \leq 5000 \\ 2400 + 0.24x & \text{if } x > 5000 \end{cases}\]

c) At \(x = 650\): 1. \(C(650) = 80 + 910 = 990\) EUR is defined 2. \(\lim_{x \to 650^-} C(x) = 990\) and \(\lim_{x \to 650^+} C(x) = 600 + 390 = 990\), so the limit exists and is 990 3. Limit = value ✓

At \(x = 5000\): 1. \(C(5000) = 600 + 3000 = 3{,}600\) EUR is defined 2. \(\lim_{x \to 5000^-} C(x) = 3{,}600\) and \(\lim_{x \to 5000^+} C(x) = 2400 + 1200 = 3{,}600\) 3. Limit = value ✓

\(C\) is continuous at both switching points, because the company switches exactly where the two methods cost the same. For a customer whose order grows slowly, the costs grow without any jump: a few more copies never make the order suddenly much more expensive.

d) Under the manager’s rule, the costs \(C_R(x)\) are \(600 + 0.6x\) below 4,000 copies and \(2400 + 0.24x\) from 4,000 copies on:

\[\lim_{x \to 4000^-} C_R(x) = 600 + 2400 = 3{,}000 \text{ EUR}, \qquad \lim_{x \to 4000^+} C_R(x) = 2400 + 960 = 3{,}360 \text{ EUR}\]

The costs jump up by 360 EUR at 4,000 copies (jump discontinuity). The rule switches too early: between 4,000 and 5,000 copies, web offset printing is still more expensive than sheet-fed offset printing.

Order of 4,500 copies: web offset \(2400 + 1080 = 3{,}480\) EUR, sheet-fed offset \(600 + 2700 = 3{,}300\) EUR. The rule costs 180 EUR too much.

e) Digital printing costs 1.40 EUR per copy for printing alone, more than the price of 1.00 EUR. Every digitally printed copy loses 0.40 EUR, and the 80 EUR for the files come on top: \(P_D(x) = x - (80 + 1.4x) = -80 - 0.4x < 0\) for every order.

Profit with the cheapest method, \(P(x) = x - C(x)\):

  • \(0 < x \leq 650\): \(P(x) = -80 - 0.4x < 0\) (loss)
  • \(650 < x \leq 5000\): \(P(x) = 0.4x - 600\), zero at \(x = 1500\)
  • \(x > 5000\): \(P(x) = 0.76x - 2400 > 0.76 \cdot 5000 - 2400 = 1400 > 0\)

\(P(x) = 0.4x - 600 > 0 \iff x > 1500\). At exactly 1,500 copies the profit is 0, so the smallest order with a profit is 1,501 copies.

Problem 10: First-Aid Courses for Companies (xx)

A training company runs first-aid courses at the sites of its customers. In its old price list, each participant pays \(90 - 2n\) EUR when \(n\) people take part in the course (a group discount). The course fee for the whole group is then \(F_0(n) = n(90 - 2n)\) EUR.

  1. Compute the course fee for 10 and for 15 participants. Determine the group size with the highest course fee in the old price list and explain why the old price list does not work for large groups.

  2. The new price list keeps the old formula for up to 15 participants. For more than 15 participants, the course fee is \(F(n) = k + 30n\) EUR. An employee suggests \(k = 300\). Determine the one-sided limits of \(F\) at \(n = 15\) and the fee for 16 participants. Explain what a company with 15 employees would do.

  3. Determine \(k\) so that \(F\) is continuous at \(n = 15\) and verify the three conditions.

(for verification: \(k = 450\))

  1. Use \(k = 450\). The course room holds at most 24 people. Compute the course fee for 24 participants and the fee per participant for 15, 20 and 24 participants. Compute by how many percent the new fee for 24 participants is higher than the fee in the old price list. Round to one decimal place.

a) \(F_0(10) = 10 \cdot 70 = 700\) EUR and \(F_0(15) = 15 \cdot 60 = 900\) EUR.

\(F_0(n) = -2n^2 + 90n\) is a parabola with \(a = -2 < 0\). It opens downward, so its vertex is a maximum:

\[n_V = -\frac{b}{2a} = -\frac{90}{2 \cdot (-2)} = 22.5, \qquad F_0(22.5) = 22.5 \cdot 45 = 1{,}012.50 \text{ EUR}\]

For whole groups: \(F_0(22) = 22 \cdot 46 = 1{,}012\) EUR and \(F_0(23) = 23 \cdot 44 = 1{,}012\) EUR. Groups of 22 or 23 participants bring the highest fee.

Beyond the vertex, the fee falls when more people join: e.g. \(F_0(30) = 30 \cdot 30 = 900\) EUR, the same as for 15 participants. The training company would teach twice as many people for the same money, and at \(n = 45\) the course would be free. The group discount must not continue for large groups.

b) With \(k = 300\):

  • \(F(15) = F_0(15) = 900\) EUR (the old formula applies up to 15)
  • \(\lim_{n \to 15^-} F(n) = 900\) EUR
  • \(\lim_{n \to 15^+} F(n) = 300 + 30 \cdot 15 = 750\) EUR
  • \(F(16) = 300 + 480 = 780\) EUR

The fee jumps down by 150 EUR at 15 participants. A company with 15 employees would register a 16th person (e.g. a trainee) and pay 780 EUR instead of 900 EUR.

c) Continuity at \(n = 15\): 1. \(F(15) = 900\) is defined 2. \(\lim_{n \to 15^-} F(n) = 900\) and \(\lim_{n \to 15^+} F(n) = k + 450\). The limit exists if \(k + 450 = 900\), so \(k = 450\) 3. Then \(\lim_{n \to 15} F(n) = 900 = F(15)\) ✓

With \(k = 450\), all three conditions hold: \(F\) is continuous at 15 participants.

d) \(F(24) = 450 + 30 \cdot 24 = 1{,}170\) EUR.

Participants Course fee Fee per participant
15 900 EUR 60.00 EUR
20 \(450 + 600 = 1{,}050\) EUR 52.50 EUR
24 1,170 EUR 48.75 EUR

Larger groups still pay less per person, but the course fee now rises with every participant.

Old price list: \(F_0(24) = 24 \cdot 42 = 1{,}008\) EUR. \(\frac{1170}{1008} \approx 1.161\), so the new fee is about 16.1 % higher.

D: FSP Problem

Work on this problem as in the exam: write down every step. The points show what each part is worth.

FSP Problem: Bicycle Trailers for Children [20 pts. total]

A young company builds bicycle trailers for children and sells them to bike shops for 450 EUR per trailer. In its own workshop, it can build at most 1,600 trailers per month (its capacity). Up to this capacity, its monthly total costs are

\[C(x) = 0.5x^2 + 20x + 150, \qquad 0 \leq x \leq 16\]

For more than 1,600 trailers per month, the company buys the additional trailers from a partner plant (another company that builds them on its behalf) for 300 EUR per trailer. For more than 2,500 trailers per month, it must also rent an additional warehouse for 40,000 EUR per month.

(QU = quantity units, MU = monetary units. 1 QU = 100 trailers, 1 MU = 1,000 EUR. Average cost = total costs divided by the output.)

Part a)

Show that the profit function for \(0 \leq x \leq 16\) is given by \(P(x) = -0.5x^2 + 25x - 150\). Determine the break-even point and give the smallest number of trailers per month with which the company makes a profit. [4 pts.]

(for verification: \(x = 25 - 5\sqrt{13}\))

Price in MU per QU: 450 EUR per trailer \(= 45{,}000\) EUR per QU \(= 45\) MU per QU, so \(R(x) = 45x\).

Profit:

\[P(x) = R(x) - C(x) = 45x - (0.5x^2 + 20x + 150) = -0.5x^2 + 25x - 150 \quad \checkmark\]

Zeros: \(-0.5x^2 + 25x - 150 = 0 \;\big|\cdot(-2) \iff x^2 - 50x + 300 = 0\). Quadratic formula with \(\Delta = 50^2 - 4 \cdot 300 = 1300\) and \(\sqrt{1300} = 10\sqrt{13}\):

\[x_{1,2} = \frac{50 \pm 10\sqrt{13}}{2} = 25 \pm 5\sqrt{13}, \qquad x_1 \approx 6.97, \qquad x_2 \approx 43.03\]

Only \(x_1\) lies in \([0; 16]\). \(P(0) = -150 < 0\) and \(P(16) = -128 + 400 - 150 = 122 > 0\), so the loss turns into a profit at \(x_1\): the break-even point is \(x_1 = 25 - 5\sqrt{13} \approx 6.972\) QU \(\approx 697.2\) trailers.

Profit from 698 trailers per month on (697 trailers still give a small loss).

Grading: 1 Pt. price 45 MU per QU and profit term, 1 Pt. zero form, 1 Pt. quadratic formula \(25 \pm 5\sqrt{13}\), 1 Pt. choice of the zero in \([0; 16]\) and 698 trailers.

Part b)

The controller writes the total costs for \(16 < x \leq 25\) as \(C(x) = 30x + k\).

Determine \(k\) so that \(C\) is continuous at \(x = 16\) and verify the three conditions for continuity. Explain why a continuous cost function makes sense here. [4 pts.]

Value at the capacity: \(C(16) = 0.5 \cdot 256 + 20 \cdot 16 + 150 = 128 + 320 + 150 = 598\) MU.

Condition for \(k\): the right-hand limit must equal the left-hand limit:

\[\lim_{x \to 16^+} C(x) = 30 \cdot 16 + k = 480 + k \overset{!}{=} 598 = \lim_{x \to 16^-} C(x) \iff k = 118\]

Three conditions with \(k = 118\): 1. \(C(16) = 598\) is defined 2. \(\lim_{x \to 16^-} C(x) = 598\) and \(\lim_{x \to 16^+} C(x) = 480 + 118 = 598\), so \(\lim_{x \to 16} C(x) = 598\) exists 3. \(\lim_{x \to 16} C(x) = 598 = C(16)\) ✓

Meaning: above the capacity, the workshop still builds 1,600 trailers at the costs \(C(16) = 598\) MU, and each further QU from the partner plant adds 30 MU (300 EUR per trailer \(= 30{,}000\) EUR per QU). So \(C(x) = 598 + 30(x - 16) = 30x + 118\). Switching to the partner plant causes no one-time costs, so there is no reason for a jump.

Grading: 1 Pt. \(C(16) = 598\), 1 Pt. condition left limit = right limit and \(k = 118\), 1 Pt. all three conditions checked, 1 Pt. explanation (1,600 own trailers plus 30 MU per further QU).

Part c)

For \(x > 25\), the total costs are \(C(x) = 30x + 158\).

Investigate the one-sided limits of \(C\) at \(x = 25\) and decide whether \(C\) is continuous there. Explain the business meaning of the result. [3 pts.]

One-sided limits:

\[\lim_{x \to 25^-} C(x) = 30 \cdot 25 + 118 = 868 \text{ MU}, \qquad \lim_{x \to 25^+} C(x) = 30 \cdot 25 + 158 = 908 \text{ MU}\]

\(C(25) = 868\) MU is defined, but the one-sided limits differ (\(868 \neq 908\)), so \(\lim_{x \to 25} C(x)\) does not exist. \(C\) is not continuous at \(x = 25\): a jump discontinuity of 40 MU.

Meaning: the jump of \(40\) MU \(= 40{,}000\) EUR is the monthly rent of the additional warehouse. As soon as the output exceeds 2,500 trailers, even by one trailer, the costs jump by the whole rent.

Grading: 1 Pt. both one-sided limits, 1 Pt. not continuous with reason (limit does not exist), 1 Pt. jump = warehouse rent 40,000 EUR.

Part d)

Give the average cost function \(AC(x)\) for \(0 < x \leq 16\). Determine \(\lim_{x \to 0^+} AC(x)\) and explain the result. [2 pts.]

\[AC(x) = \frac{C(x)}{x} = \frac{0.5x^2 + 20x + 150}{x} = 0.5x + 20 + \frac{150}{x}, \qquad 0 < x \leq 16\]

For \(x \to 0^+\), \(0.5x + 20 \to 20\), but \(\frac{150}{x}\) grows without bound:

\[\lim_{x \to 0^+} AC(x) = +\infty\]

Explanation: the fixed costs of \(150\) MU \(= 150{,}000\) EUR per month arise even without production. With very few trailers, they are spread over very few trailers, so each trailer becomes extremely expensive.

Grading: 1 Pt. termwise division, 1 Pt. limit \(+\infty\) with the fixed costs as the reason.

Part e)

For \(x > 25\), the average cost is \(AC(x) = 30 + \frac{158}{x}\).

Compute the limit of \(AC(x)\) at positive infinity and interpret its significance. [3 pts.]

Limit: \(\frac{158}{x} \to 0\) for \(x \to \infty\), so

\[\lim_{x \to \infty} AC(x) = \lim_{x \to \infty} \left(30 + \frac{158}{x}\right) = 30 \text{ MU per QU}\]

Conversion: \(30\) MU per QU \(= \frac{30{,}000 \text{ EUR}}{100 \text{ trailers}} = 300\) EUR per trailer.

Interpretation: for very large outputs, the average cost comes closer and closer to 300 EUR per trailer, the price of the partner plant. The constant part of \(158\) MU does not grow with the output. It is spread over more and more trailers, and its share per trailer goes to 0. Since \(\frac{158}{x} > 0\), the average cost always stays above 300 EUR. With a price of 450 EUR, the profit per trailer approaches 150 EUR, but never reaches it.

Grading: 1 Pt. limit 30 MU per QU with \(\frac{158}{x} \to 0\), 1 Pt. conversion to 300 EUR per trailer, 1 Pt. interpretation (partner price as the lower bound, fixed part spread over many trailers).

Part f)

Compute the profit for 2,400, 2,500 and 2,600 trailers per month. Decide whether the company should increase its output from 2,500 to 2,600 trailers. Determine the smallest number of trailers above 2,500 with which the profit is higher than at 2,500 trailers. [4 pts.]

Profit on the pieces with \(R(x) = 45x\):

  • \(16 < x \leq 25\): \(P(x) = 45x - (30x + 118) = 15x - 118\)
  • \(x > 25\): \(P(x) = 45x - (30x + 158) = 15x - 158\)

Values:

Trailers \(x\) in QU Profit in MU Profit in EUR
2,400 24 \(360 - 118 = 242\) 242,000
2,500 25 \(375 - 118 = 257\) 257,000
2,600 26 \(390 - 158 = 232\) 232,000

Decision: from 2,500 to 2,600 trailers, the profit falls by 25 MU \(= 25{,}000\) EUR. The extra 100 trailers bring only \(15\) MU, but the warehouse costs \(40\) MU. The company should not increase its output to 2,600 trailers.

Threshold: for \(x > 25\), the profit must exceed 257 MU:

\[15x - 158 > 257 \iff 15x > 415 \iff x > 27.67\]

\(27.67\) QU \(\approx 2{,}766.7\) trailers. The profit is higher than at 2,500 trailers from 2,767 trailers per month on.

Grading: 1 Pt. profit terms on both pieces, 1 Pt. the three profits, 1 Pt. decision with the reason (warehouse rent larger than the extra contribution), 1 Pt. inequality and 2,767 trailers.