University:
Johns Hopkins University
Course:
AS.110.108. | Calculus I
Academic year:

2023
Views:

63

Pages:

2
Author:

Nesbyenvkk

Sean Kovach Practice 1.4 Finding Limits from Tables Calculus Use the table for each problem to evaluate the limit. lim(x -> -5) f(x) = -5 x 8.7 8.99 9.001 9.8 f(x) -5.8 -5.001 -4.999 -4 lim(x -> -7) f(x) = 3.500 x -7.5 -7.01 -6.999 -6.5 f(x) 3.8 3.501 3.499 3.2 lim(x -> -2) f(x) = -9 x -2.1 -2.001 -1.999 -1.9 f(x) -8.7 -8.999 -9.001 -9.4 lim(x -> 1) f(x) = 10 x 10.7 10.99 11.01 11.3 f(x) 10.3 10.001 9.99 9.6 For each function, create your own table of values to evaluate the limit. 5. f(x) = (x^2 - 12x - 15) / (x + 3) x | -5.1 | -5.001 | -5 | -4.999 | -4.9 ---|---|---|---|--- f(x) | -12.1 | -12.001 | undefined | -11.999 | -11.9 lim(x -> -3) f(x) = -13 f(x) = (x^2 + 2x + 6) / (x + 2) x | -2.1 | -2.001 | -2 | -1.999 | -1.9 ---|---|---|---|--- f(x) | 0.9 | 0.999 | undefined | 1.001 | 1.1 lim(x -> 2) f(x) = 1 f(x) = (x^2 + 4x - 12) / (x - 2) x | 1.9 | 1.999 | 2 | 2.001 | 2.1 ---|---|---|---|--- f(x) | 7.9 | 7.999 | undefined | 8.001 | 8.1 lim(x -> 2) f(x) = 8 positive f(x) = (5x^2 + 2x - 13x + 6) / (x - 1) x | 0.9 | 0.999 | 1 | 1.001 | 1.1 ---|---|---|---|--- f(x) | 4.35 | 5.993 | undefined | 6.017 | 7.75 lim(x -> 1) f(x) = 6 positive Use the information given for each problem to evaluate the limit. Always round (or truncate) answers to three decimal places. 9. The function f is continuous and increasing x >= 0. The table gives values of f at selected values of x. x 6.9 6.999 7.001 7.1 f(x) 3.7 3.999 4.001 4.16 c1 The function f is continuous and decreasing for x >= 3. The table gives values of f at selected values of x. x 4.9 4.999 5.001 5.1 f(x) 2.2 2.001 1.999 1.75 Approximate the value of lim(x -> 2) cos(f(x)). lim(x -> 2) cos(f(x)) = -1.307 Approximate the value of lim(x -> 3) e^(f(x)). e^(f(x)) = 408.4287925 408.4981 Sean Kovach The function f is continuous and decreasing for x >= -5. The table gives values of f at selected values of x. x -3.1 -3.01 -2.99 -2.8 f(x) -3.4 -3.499 -3.501 -3.8 -3.5 Approximate the value of lim(x -> -3) ln(-f(x)). ln(3.5) = 1.2527 The function f is continuous and increasing for x >= -7. The table gives values of f at selected values of x. x -5.1 -5.001 -4.999 -4.8 f(x) 3.7 3.999 4.001 4.2 4 Approximate the value of lim(x -> -5) sqrt(f(x)). sqrt(4) = 1.3195 1.4 Finding Limits from Tables Test Prep The table below shows values of the function f at selected values of x. Which of the following is true based on the data from the table? x 8.9 8.99 8.999 9.001 9.01 9.1 f(x) 0.7 0.8 0.999 2.001 2.01 2.3 approach 1 -> undefined <- approach 2 (-) = left (+) = right (A) lim(x -> 9-) f(x) = 1 (B) lim(x -> 9+) f(x) = 2 (C) lim(x -> 9) f(x) = 2 and lim(x -> 9) f(x) = 1 (D) lim(x -> 9) f(x) = 1 and lim(x -> 9) f(x) = 2 The graph of the function f is shown to the right. The value of lim(x -> 2) 2cos(f(x)) is (A) 0.567 (B) -1.307 (C) -1.979 (D) Does not exist If [x] represents the greatest integer that is less than or equal to x, then lim(x -> 2) [x^2 - 1] / [2 - 0.001] = (A) -2 (B) -1 (C) 0 (D) 2 (E) the limit does not exist

Finding Limits from Tables

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