Math, Limits and Continuity Worksheet

Question 1

Evaluate the limit: lim as x approaches 3 of (2x + 5).

Accepted Answer

The limit is 11 because substituting x = 3 into the linear expression gives 2(3) + 5 = 11.

Question 2

Evaluate the limit: lim as x approaches -2 of (x^2 - 4x + 1).

Accepted Answer

The limit is 13 because polynomials are continuous, so direct substitution gives (-2)^2 - 4(-2) + 1 = 4 + 8 + 1 = 13.

Question 3

Evaluate the limit: lim as x approaches 4 of (x^2 - 16) / (x - 4).

Accepted Answer

The limit is 8 because x^2 - 16 factors as (x - 4)(x + 4). After canceling x - 4, the expression becomes x + 4, and substituting x = 4 gives 8.

Question 4

Evaluate the limit: lim as x approaches 2 of (x^2 + x - 6) / (x - 2).

Accepted Answer

The limit is 5 because x^2 + x - 6 factors as (x + 3)(x - 2). After canceling x - 2, the expression becomes x + 3, and substituting x = 2 gives 5.

Question 5

Evaluate the limit: lim as x approaches 0 of sin(x) / x.

Accepted Answer

The limit is 1. This is a standard trigonometric limit that holds when x is measured in radians.

Question 6

Evaluate the limit: lim as x approaches infinity of 5 / x.

Accepted Answer

The limit is 0 because as x becomes larger and larger, the fraction 5 divided by x gets closer and closer to 0.

Question 7

Evaluate the limit: lim as x approaches infinity of (3x^2 + 1) / (x^2 - 4).

Accepted Answer

The limit is 3 because the highest-degree terms dominate for large x. Dividing numerator and denominator by x^2 gives (3 + 1/x^2) / (1 - 4/x^2), which approaches 3/1 = 3.

Question 8

Find the one-sided limits for f(x) = |x| / x at x = 0. State the left-hand limit and the right-hand limit, then say whether the two-sided limit exists.

Accepted Answer

As x approaches 0 from the left, |x| / x = -1, so the left-hand limit is -1. As x approaches 0 from the right, |x| / x = 1, so the right-hand limit is 1. Because these one-sided limits are not equal, the two-sided limit does not exist.

Question 9

A function is defined by f(x) = (x^2 - 1) / (x - 1) for x not equal to 1, and f(1) = 5. Is the function continuous at x = 1? Explain.

Accepted Answer

The function is not continuous at x = 1. For x not equal to 1, the expression simplifies to x + 1, so the limit as x approaches 1 is 2. Since f(1) = 5 and the function value does not match the limit, the function is not continuous at x = 1.

Question 10

A function is defined by f(x) = x^2 for x less than or equal to 1, and f(x) = 2x - 1 for x greater than 1. Is f(x) continuous at x = 1? Explain.

Accepted Answer

Yes, the function is continuous at x = 1. From the left, x^2 approaches 1^2 = 1. From the right, 2x - 1 approaches 2(1) - 1 = 1. Also, f(1) = 1, so the left-hand limit, right-hand limit, and function value are all equal.

Question 11

Use the table values to estimate lim as x approaches 2 of f(x): when x = 1.9, f(x) = 3.8; when x = 1.99, f(x) = 3.98; when x = 2.01, f(x) = 4.02; when x = 2.1, f(x) = 4.2.

Accepted Answer

The estimated limit is 4 because the function values from both sides of x = 2 are getting closer to 4.

Question 12

Determine whether the function f(x) = 1 / (x - 6) is continuous at x = 6. Explain.

Accepted Answer

The function is not continuous at x = 6 because the denominator is 0 there, so the function is undefined. This creates a vertical asymptote at x = 6.

Question 13

Evaluate the limit: lim as x approaches -1 of (x^3 + 1) / (x + 1).

Accepted Answer

The limit is 3 because x^3 + 1 factors as (x + 1)(x^2 - x + 1). After canceling x + 1, substitute x = -1 into x^2 - x + 1 to get 1 - (-1) + 1 = 3.

Question 14

State whether each function is continuous everywhere on the real numbers: f(x) = 4x - 7, g(x) = x^2 + 3x + 1, and h(x) = 1 / (x + 2).

Accepted Answer

The function f(x) = 4x - 7 is continuous everywhere because linear functions are continuous for all real numbers. The function g(x) = x^2 + 3x + 1 is continuous everywhere because polynomials are continuous for all real numbers. The function h(x) = 1 / (x + 2) is not continuous everywhere because it is undefined at x = -2.

Question 15

A graph has an open circle at (3, 7) and a filled point at (3, 4). The curve approaches y = 7 from both sides as x approaches 3. Find lim as x approaches 3 of f(x), and state f(3).

Accepted Answer

The limit as x approaches 3 is 7 because the graph approaches y = 7 from both sides. The value f(3) is 4 because the filled point at x = 3 is located at y = 4.

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