Limits & Continuity - IB Mathematics Applications and Interpretation

1. Calculate the value of $\lim_{h \to 2^+} C(h) - \lim_{h \to 2^-} C(h)$.
   Answer: 3 First, we find the limit from the right, $\lim_{h \to 2^+} C(h)$. For values of $h$ slightly greater than 2 (e.g., 2.1 hours), the car is in its third hour of parking. The cost for $2 < h \le 3$ would be $8+3=11$. So, $\lim_{h \to 2^+} C(h) = 11$. Second, we find the limit from the left, $\lim_{h \to 2^-} C(h)$. For values of $h$ slightly less than 2 (e.g., 1.9 hours), the car is in its second hour of parking, so the cost is $8$. Thus, $\lim_{h \to 2^-} C(h) = 8$. The difference is $11 - 8 = 3$.
2. According to the model, what is the carrying capacity of the lake for this species of fish? This is the value the population approaches in the long run.
   Answer: 4000 The carrying capacity is the limit of the population function as time $t$ approaches infinity. For the rational function $P(t) = \frac{4000t + 500}{t+2}$, the degree of the numerator and the denominator are both 1. The limit is the ratio of the leading coefficients, which is $\frac{4000}{1} = 4000$.
3. Which statement correctly describes the population model at $t=6$?
   Answer: The model has $\lim_{t \to 6^-} P(t) = 26$ and $P(6)=24$. The limit from the left, $\lim_{t \to 6^-} P(t)$, represents the population just before the predator is introduced. Using the original model, this is $0.5(6^2) + 8 = 0.5(36) + 8 = 18 + 8 = 26$ thousand. The passage states that the measured population at $t=6$ (i.e., just after the event) is 24 thousand, so $P(6)=24$. Because the limit from the left (26) does not equal the function's value (24), the function is discontinuous. Statement C correctly identifies both values.
4. At which data usage level(s) is the cost function $C(g)$ discontinuous?
   Answer: At $g=10$ only We check the boundary points. At $g=2$: $C(2)=25$. $\lim_{g \to 2^-} C(g) = 25$. $\lim_{g \to 2^+} C(g) = 25 + 15(2-2) = 25$. Since the value and both one-sided limits are 25, the function is continuous at $g=2$. At $g=10$: $C(10) = 25 + 15(10-2) = 25 + 15(8) = 25+120=145$. The limit from the left is $\lim_{g \to 10^-} C(g) = 145$. The limit from the right is $\lim_{g \to 10^+} C(g) = 150$. Since the left-hand limit (145) does not equal the right-hand limit (150), the function is discontinuous at $g=10$.
5. What temperature does the coffee approach as it is left to cool for a very long time?
   Answer: 22°C We need to find the limit of the temperature function as $t \to \infty$. As $t \to \infty$, the term $e^{-0.04t}$ approaches 0. Therefore, $\lim_{t \to \infty} T(t) = \lim_{t \to \infty} (22 + 68e^{-0.04t}) = 22 + 68(0) = 22$. The coffee's temperature approaches the room temperature.
6. What is the physical interpretation of the mathematical expression $\lim_{h \to 0} \frac{s(5+h)-s(5)}{h}$?
   Answer: The instantaneous speed of the cyclist at exactly 5 seconds. The expression $\frac{s(5+h)-s(5)}{h}$ represents the average rate of change of distance (average speed) over a small time interval $h$ starting at $t=5$. The limit as this interval $h$ approaches zero gives the instantaneous rate of change of distance at $t=5$. The instantaneous rate of change of distance is the definition of instantaneous speed (or velocity).
7. Based on the numerical data, what is the best estimate for $\lim_{x \to 3} R(x)$?
   Answer: 15.0 When you encounter a table of values approaching a specific point, you're being asked to estimate a limit using numerical evidence. This tests your understanding of limits as the behavior of a function as the input approaches a particular value. To find $\lim_{x \to 3} R(x)$, examine what happens to the function values as $x$ gets closer and closer to 3 from both sides. Looking at the table, as $x$ approaches 3 from the left (2.9 → 2.99 → 2.999), $R(x)$ approaches 15: the values are 14.71, 14.970, and 14.997. From the right (3.001 ← 3.01 ← 3.1), $R(x)$ also approaches 15: the values are 15.003, 15.030, and 15.31. The values closest to $x = 3$ (at 2.999 and 3.001) give $R(x)$ values of 14.997 and 15.003, both extremely close to 15. Answer A (15.0) is correct because both one-sided limits approach this value, indicating the limit exists and equals 15. Answer B (14.997) is wrong because this is just one specific function value, not the limit that both sides approach. Answer C (14.71) is incorrect because this value occurs at $x = 2.9$, which is relatively far from 3 and doesn't represent the limiting behavior. Answer D is wrong because the limit clearly exists—both sides approach the same value. Study tip: For numerical limit problems, always check values from both sides of the target point. The limit exists when both sides approach the same number, regardless of what happens exactly at that point.
8. Which statement best describes the balloon's altitude as time $t$ gets extremely close to 10 seconds?
   Answer: The altitude oscillates between 496 and 504 metres with increasing frequency. When analyzing functions with extreme behavior near critical points, you need to examine what happens to each component as the variable approaches that value. Here, as $t$ approaches 10, the expression $\frac{1}{t-10}$ becomes increasingly large in magnitude (approaching $\pm\infty$). The key insight is understanding how $\sin(\frac{1}{t-10})$ behaves as $t \to 10$. Since the argument $\frac{1}{t-10}$ grows without bound, the sine function oscillates between -1 and +1 with increasing frequency. This means $4\sin(\frac{1}{t-10})$ oscillates between -4 and +4, making the complete function $A(t) = 500 + 4\sin(\frac{1}{t-10})$ oscillate between 496 and 504 metres. Option A is wrong because the altitude doesn't stabilize at 500 metres—it continuously oscillates around this value. Option B incorrectly suggests the altitude approaches 504 metres, which would only occur if the sine function approached +1, but it oscillates between all values. Option C is incorrect because while the frequency increases toward infinity, the altitude itself remains bounded between 496 and 504 metres due to the sine function's limited range. Option D correctly captures both essential behaviors: the oscillation between the bounded values (496 to 504 metres) and the increasing frequency as $t$ approaches 10. Study tip: When you see $\frac{1}{x-a}$ inside a trigonometric function near $x = a$, immediately think "oscillating with increasing frequency" rather than approaching a single limit value.
9. How is the discontinuity in the price function $P(t)$ at midnight best classified?
   Answer: A jump discontinuity When analyzing functions that represent real-world scenarios with sudden changes, you need to identify what type of discontinuity occurs at the point of change. Discontinuities are classified based on the behavior of the function's left and right limits. At midnight, the subscription price jumps instantly from $10 to $12. This creates a situation where the left-hand limit (approaching midnight from before) is $\lim_{t \to 0^-} P(t) = 10$, while the right-hand limit (approaching midnight from after) is $\lim_{t \to 0^+} P(t) = 12$. Since these one-sided limits exist but are not equal, this is a jump discontinuity, making D correct. Let's examine why the other options don't apply. Option A (removable discontinuity) occurs when the left and right limits exist and are equal, but the function value at that point is either undefined or different from the limit—you could "remove" the discontinuity by redefining the function at that single point. That's not the case here since the limits aren't equal. Option B (infinite discontinuity) happens when at least one of the one-sided limits approaches infinity, which clearly doesn't occur with prices jumping from $10 to $12. Option C (continuous) would require the left limit, right limit, and function value to all be equal at midnight, but we have a clear price jump. For IB exam success, remember that jump discontinuities are common in piecewise functions representing real-world situations like pricing changes, tax brackets, or shipping rates—look for sudden "jumps" between finite values.
10. What is the interpretation of $\lim_{t \to \infty} P(t)$ in the context of this model?
    Answer: The employee will eventually achieve a maximum proficiency of 100%. We need to evaluate the limit of $P(t)$ as $t \to \infty$. Since the base of the exponential term is 0.85 (which is between 0 and 1), the term $(0.85)^t$ approaches 0 as $t$ becomes very large. Therefore, $\lim_{t \to \infty} P(t) = 100 - 80(0) = 100$. This means that with enough training, the model predicts the employee's proficiency will approach a maximum of 100%.
11. Which statement correctly interprets the meaning of $\lim_{n \to 5^+} E(n)$?
    Answer: The effort to memorize slightly more than 5 items becomes infinitely large. We are evaluating the limit as $n$ approaches 5 from the right (positive) side. As $n \to 5^+$, the numerator $2n^2$ approaches $2(5^2) = 50$. The denominator $n-5$ approaches 0 from the positive side (e.g., 5.01 - 5 = 0.01). A positive constant divided by a very small positive number results in a very large positive number. Therefore, the limit is $+\infty$. This suggests the model is not valid at or very near $n=5$, as it predicts an infinite effort.
12. Which statement correctly describes the continuity of the velocity model?
    Answer: The model is continuous on the interval $0 \le t \le 60$. When you encounter piecewise functions in IB Mathematics, the key question is whether the pieces "connect" smoothly at their boundary points. Continuity means there are no jumps, gaps, or breaks in the function. To determine continuity at $t = 50$, you need to check if the left-hand limit equals the right-hand limit at this point. From the left (using $v(t) = 0.1t^2$): $v(50) = 0.1(50)^2 = 250$ km/s. From the right (using $v(t) = At + B$), the velocity approaches $50A + B$. Here's the crucial insight: since the problem states that $v(t)$ is continuous for the entire 60-second interval, the constants $A$ and $B$ must be chosen specifically to ensure $50A + B = 250$. The problem wouldn't present an inconsistent model. Answer A is wrong because physical events (like an engine turning off) don't automatically create mathematical discontinuity. The velocity itself can remain continuous even when acceleration changes. Answer B reflects a common misconception that changing formulas means discontinuity. Piecewise functions can absolutely be continuous if their pieces align properly at transition points. Answer C suggests we need to know the specific values of $A$ and $B$, but the problem statement already tells us the function is continuous, which constrains these values appropriately. Answer D is correct because the problem explicitly states the function is continuous on the given interval. Study tip: In IB exam questions, when a problem describes a physical model as continuous, trust that statement—it's telling you the mathematical pieces fit together properly, even if you don't see all the specific values.
13. To create a continuous velocity model, what value should be assigned to $v(3)$?
    Answer: 12 m/s For continuity at $t=3$, we need $v(3)$ to equal the limit as $t$ approaches 3. We can simplify the expression by factoring: $v(t) = \frac{2(t^2 - 9)}{t-3} = \frac{2(t-3)(t+3)}{t-3} = 2(t+3)$ for $t \ne 3$. Therefore, $\lim_{t \to 3} v(t) = 2(3+3) = 12$. Setting $v(3) = 12$ ensures the velocity model is continuous.
14. Which of the following financial events would be best modelled by a jump discontinuity in the function $B(t)$?
    Answer: The bank processing a deposited cheque at a specific time. A jump discontinuity represents an instantaneous change in value. A cheque being deposited and processed causes the account balance to jump up instantly at a specific point in time. Continuous interest (A) results in a smooth, continuous increase. A forecast (C) is a prediction, not a change in the actual balance function. Inflation (D) is a gradual, continuous process.
15. What does the model predict for the long-term concentration of the drug in the bloodstream as time increases indefinitely?
    Answer: 0 mg/L To find the long-term concentration, we evaluate the limit as $t \to \infty$. For the rational function $C(t) = \frac{120t}{t^2 + 4}$, the degree of the denominator (2) is greater than the degree of the numerator (1). Therefore, the limit as $t$ approaches infinity is 0. This means the drug concentration approaches zero as the drug is eventually eliminated from the patient's system.
16. Which statement correctly interprets the value of $\lim_{x \to 0^+} AC(x)$?
    Answer: As production begins, the average cost per unit becomes infinitely large. We are evaluating the limit of $AC(x)$ as $x$ approaches 0 from the positive side. As $x \to 0^+$, the numerator $12x + 4800$ approaches 4800, and the denominator $x$ approaches 0. A constant divided by a very small positive number results in a very large positive number. Thus, $\lim_{x \to 0^+} AC(x) = \infty$. This means the fixed costs are spread over very few units, making the average cost per unit extremely high.
17. What must be the value of the constant $k$ for the cost function $C(v)$ to be continuous at $v=20$?
    Answer: 10 For the function to be continuous at $v=20$, the limit from the left must equal the limit from the right. We set the two expressions equal to each other at $v=20$: $\lim_{v \to 20^-} C(v) = \lim_{v \to 20^+} C(v)$. This gives $2(20) = k + 1.5(20)$. So, $40 = k + 30$. Solving for $k$ gives $k = 40 - 30 = 10$.

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