Estimating Limit Values from Tables: Practice Questions (Calculus AB 1.4)

Answer: (B). From the left, 2.41, 2.491, 2.4991 approach 2.5. From the right, 2.59, 2.509, 2.5009 also approach 2.5. Both sides agree, so the limit is about 2.5. The entry f(4) = 7 does not affect the limit.

• (A) is just the table value closest to x = 4 on the left. The values are approaching 2.5, so 2.5 is the better estimate.
• (C) uses f(4). A limit depends on values near 4, not at 4.
• (D) assumes that f(4) differing from the trend destroys the limit. It does not; it only means f is not continuous at 4 (Topic 1.10 onwards).

Answer: (A). “x → −1⁺” means from the right, so use x values greater than −1: −0.9, −0.99, −0.999. The values 0.7, 0.97, 0.997 approach 1.

• (B) averages the left-side trend (3) and the right-side trend (1). Limits are never found by averaging.
• (C) is the left-hand limit. Values such as −1.01 are less than −1, so they are on the left.
• (D) is true of the two-sided limit, because 3 ≠ 1. But the question asks only for the right-hand limit, which does exist.

Answer: (C). Each step closer to 5 multiplies the values by 100: 50, 5 000, 500 000. They do not level off, so there is no finite number they approach. The function is unbounded near 5, which is one of the ways a limit fails to exist (you may write lim (x → 5) f(x) = ∞ to describe it).

• (A) takes the largest values in the table as the limit. The next step closer would give 50 000 000.
• (B) may come from noticing that x − 5 gets close to 0. The question is about f(x), which gets large, not small.
• (D) confuses “the two sides match each other” with “the two sides approach the same real number”. Matching values that grow without bound still give no finite limit.

Answer: (B). From the left the values rise towards a number just above 8.041. From the right they fall towards a number just below 8.054. The limit lies between 8.041 and 8.054, and 8.05 is the only option in that interval.

• (A) is f(0.9), the value furthest from x = 1 on the left.
• (C) is f(1.1), the value furthest from x = 1 on the right.
• (D) confuses the limit with the function value. The limit can exist even when f(1) does not, as here (substitution would give 0/0).

Background only: the exact value is 5 ln 5 ≈ 8.047, which you will be able to show once you can differentiate exponential functions, later in the course.

(a) f(1/11) = cos(11π) = −1, because 11π is an odd multiple of π. Likewise f(1/101) = cos(101π) = −1.

(b) For every x in the table, π/x is an even multiple of π (for example π/0.1 = 10π and π/0.5 = 2π), and the cosine of an even multiple of π is 1. The table sampled only the points where f happens to equal 1. Other x values, just as close to 0 or closer, give −1.

(c) No. However close x gets to 0, there are x values with f(x) = 1 (x = 1/(2n)) and x values with f(x) = −1 (x = 1/(2n + 1)), for whole numbers n. The values oscillate between −1 and 1 and never approach a single number, so the limit does not exist.

Lesson: a table is a sample. A pattern that looks perfect can be produced by the choice of x values.

Suggested Marlbridge rubric (3 points): 1 for both values −1 in (a); 1 for explaining that every table value makes π/x an even multiple of π; 1 for “does not exist” with the oscillation reason (values 1 and −1 occur arbitrarily close to 0).

(a) From the left (h < 0): 5.705, 5.97005, 5.9970005 approach 6. From the right (h > 0): 6.305, 6.03005, 6.0030005 also approach 6. So lim (h → 0) A(h) ≈ 6 m/s.

(b) Average velocity is (change in position)/(change in time). When h = 0 the time interval has length 0, so the calculation would divide by 0. A(0) is undefined. That does not matter: the limit only uses h values near 0.

(c) A two-sided limit exists only if the left-hand and right-hand limits are equal. The negative values show the left-hand trend and the positive values show the right-hand trend. Here both approach 6, so the two-sided limit exists.

(d) The average velocity over shorter and shorter time intervals around t = 2 approaches 6 m/s. So the robot’s velocity at the instant t = 2 seconds is about 6 m/s. (This is the idea of Topic 1.1: an instantaneous rate is a limit of average rates.)

Acceptable alternative for (a): “approximately 6.0 m/s”. An answer of 5.9970005 or 6.0030005 alone does not earn the first point; it is a table value, not the value being approached.

(a)

(b) The left values increase towards a number above 0.6924. The right values decrease towards a number below 0.6939. The two closest values, 0.6924 and 0.6939, both round to 0.69 but give 0.692 and 0.694 to 3 decimal places. So lim (x → 0) h(x) ≈ 0.69, to 2 decimal places. (Going closer, to x = ±0.0001, would support a third decimal place.)

(c) A table shows only a few sampled values. It cannot show what happens for x values between the samples or closer to 0. The function could, in principle, behave differently there, as cos(π/x) does in Question 5. A table is evidence for a limit, not proof.

Background only: the exact value is ln 2 ≈ 0.6931, which you will be able to show once you can differentiate exponential functions, later in the course.