Calculus: Limits and Continuity Worksheet

Question 1

Evaluate the limit: lim as x approaches 3 of (x^2 - 9)/(x - 3).

Accepted Answer

The limit is 6. Factoring gives (x - 3)(x + 3)/(x - 3), which simplifies to x + 3 for x not equal to 3, so the limit is 3 + 3 = 6.

Question 2

Evaluate the limit: lim as x approaches 0 of sin(5x)/x.

Accepted Answer

The limit is 5. Rewrite sin(5x)/x as 5 times sin(5x)/(5x), and use the standard limit lim as u approaches 0 of sin(u)/u = 1.

Question 3

Evaluate the limit: lim as x approaches 2 of (x^3 - 8)/(x - 2).

Accepted Answer

The limit is 12. Factoring the difference of cubes gives (x - 2)(x^2 + 2x + 4)/(x - 2), so the expression simplifies to x^2 + 2x + 4, and substituting x = 2 gives 4 + 4 + 4 = 12.

Question 4

Evaluate the one-sided limits for f(x) = |x|/x as x approaches 0 from the left and from the right. Then state whether lim as x approaches 0 of f(x) exists.

Accepted Answer

As x approaches 0 from the left, |x|/x = -1. As x approaches 0 from the right, |x|/x = 1. Since the one-sided limits are different, the two-sided limit does not exist.

Question 5

Determine whether the function f(x) = (x^2 - 4)/(x - 2) is continuous at x = 2 if f(2) is defined to be 5.

Accepted Answer

The function is not continuous at x = 2. The limit as x approaches 2 is 4 because the expression simplifies to x + 2 for x not equal to 2, but f(2) = 5, so the limit does not equal the function value.

Question 6

Find the value of k that makes the piecewise function continuous at x = 1: f(x) = x^2 + k for x less than 1, and f(x) = 3x + 1 for x greater than or equal to 1.

Accepted Answer

The value is k = 3. The left-hand limit at x = 1 is 1^2 + k = 1 + k, and the right-hand value is 3(1) + 1 = 4, so 1 + k = 4 and k = 3.

Question 7

Evaluate the limit: lim as x approaches infinity of (4x^2 - 3x + 1)/(2x^2 + 5).

Accepted Answer

The limit is 2. For a rational function with the same highest degree in the numerator and denominator, the limit at infinity is the ratio of the leading coefficients, which is 4/2 = 2.

Question 8

Evaluate the limit: lim as x approaches infinity of (7x - 2)/(x^2 + 1).

Accepted Answer

The limit is 0. The denominator has degree 2 while the numerator has degree 1, so the denominator grows faster than the numerator as x becomes large.

Question 9

A graph of f has an open circle at (2, 4), a filled point at (2, 1), and the curve approaches y = 4 from both sides of x = 2. Find lim as x approaches 2 of f(x), f(2), and state whether f is continuous at x = 2.

Accepted Answer

The limit as x approaches 2 is 4 because the graph approaches y = 4 from both sides. The function value is f(2) = 1 because the filled point is at (2, 1). The function is not continuous at x = 2 because the limit and function value are not equal.

Question 10

Use direct substitution to evaluate lim as x approaches -1 of (2x^3 - x + 4).

Accepted Answer

The limit is 3. Polynomial functions are continuous everywhere, so substituting x = -1 gives 2(-1)^3 - (-1) + 4 = -2 + 1 + 4 = 3.

Question 11

Evaluate the limit: lim as x approaches 0 of (1 - cos x)/x.

Accepted Answer

The limit is 0. Multiplying by (1 + cos x)/(1 + cos x) gives sin^2 x divided by x(1 + cos x), which can be written as (sin x/x)(sin x/(1 + cos x)); this approaches 1 times 0/2 = 0.

Question 12

Classify the discontinuity at x = 3 for f(x) = (x + 1)/(x - 3).

Accepted Answer

The function has an infinite discontinuity at x = 3. The denominator is zero at x = 3 and the numerator is not zero, so there is a vertical asymptote at x = 3.

Question 13

Evaluate lim as x approaches 4 of (sqrt(x) - 2)/(x - 4).

Accepted Answer

The limit is 1/4. Rationalizing the numerator gives 1/((sqrt(x) + 2)), and substituting x = 4 gives 1/(2 + 2) = 1/4.

Question 14

Determine whether the Intermediate Value Theorem guarantees a solution to x^3 + x - 1 = 0 on the interval [0, 1].

Accepted Answer

Yes, the Intermediate Value Theorem guarantees a solution. The function f(x) = x^3 + x - 1 is continuous on [0, 1], f(0) = -1, and f(1) = 1, so the function changes sign and must equal 0 somewhere between 0 and 1.

Question 15

For the piecewise function f(x) = 2x + 1 for x less than 0, f(x) = k for x = 0, and f(x) = x^2 + 1 for x greater than 0, find k so that f is continuous at x = 0.

Accepted Answer

The value is k = 1. The left-hand limit at 0 is 2(0) + 1 = 1, and the right-hand limit at 0 is 0^2 + 1 = 1, so the two-sided limit is 1. For continuity, f(0) must equal 1, so k = 1.

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Calculus: Limits and Continuity

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