Infinite Limits & Vertical Asymptotes Quiz

1) Evaluate lim_x→2⁺ 1/(x-2)

0

1/2

+∞

-∞

As x approaches 2 from the right (x→2⁺), the denominator (x-2) approaches 0 from the positive side. When we have 1 divided by a very small positive number, the result becomes very large and positive. Therefore, the limit is positive infinity.

Explanation

As x approaches 2 from the right (x→2⁺), the denominator (x-2) approaches 0 from the positive side. When we have 1 divided by a very small positive number, the result becomes very large and positive. Therefore, the limit is positive infinity.

2) Evaluate lim_{x→3⁻} 5/(3-x)

0

5/3

+∞

-∞

As x approaches 3 from the left (x→3⁻), x is less than 3, so (3-x) is positive and approaches 0. When we divide 5 by a very small positive number, the result becomes very large and positive.

Explanation

As x approaches 3 from the left (x→3⁻), x is less than 3, so (3-x) is positive and approaches 0. When we divide 5 by a very small positive number, the result becomes very large and positive.

3) Which of the following functions has a vertical asymptote at x = 1?

F(x) = (x-1)/(x+1)

F(x) = x/(x+1)

F(x) = (x+1)/(x-1)

F(x) = 1/(x²+1)

A vertical asymptote occurs where the denominator equals zero and the numerator is not zero at that point. For f(x) = (x+1)/(x-1), the denominator equals zero when x = 1, and the numerator equals 2 at x = 1, which is not zero. Therefore, there is a vertical asymptote at x = 1.

Explanation

A vertical asymptote occurs where the denominator equals zero and the numerator is not zero at that point. For f(x) = (x+1)/(x-1), the denominator equals zero when x = 1, and the numerator equals 2 at x = 1, which is not zero. Therefore, there is a vertical asymptote at x = 1.

4) Evaluate lim_{x→0⁺} 1/x²

0

1

+∞

-∞

As x approaches 0 from the right (x→0⁺), x² approaches 0 from the positive side. When we have 1 divided by a very small positive number squared, the result becomes very large and positive. Since x² is always positive for x ≠ 0, the limit is positive infinity.

Explanation

As x approaches 0 from the right (x→0⁺), x² approaches 0 from the positive side. When we have 1 divided by a very small positive number squared, the result becomes very large and positive. Since x² is always positive for x ≠ 0, the limit is positive infinity.

5) For the function f(x) = 1/(x²-4), what is the behavior as x approaches 2 from the right?

F(x) approaches 0

F(x) approaches 1/4

F(x) approaches +∞

F(x) approaches -∞

As x approaches 2 from the right (x→2⁺), the denominator (x²-4) approaches 0 from the positive side because (2)²-4 = 0 and for x slightly greater than 2, x² is slightly greater than 4. When we divide 1 by a very small positive number, the result becomes very large and positive.

Explanation

As x approaches 2 from the right (x→2⁺), the denominator (x²-4) approaches 0 from the positive side because (2)²-4 = 0 and for x slightly greater than 2, x² is slightly greater than 4. When we divide 1 by a very small positive number, the result becomes very large and positive.

6) Evaluate lim_{x→π⁺} 1/(x-π)

0

1

+∞

-∞

As x approaches π from the right (x→π⁺), the denominator (x-π) approaches 0 from the positive side. When we have 1 divided by a very small positive number, the result becomes very large and positive.

Explanation

As x approaches π from the right (x→π⁺), the denominator (x-π) approaches 0 from the positive side. When we have 1 divided by a very small positive number, the result becomes very large and positive.

7) Which statement best describes infinite limits?

A function has an infinite limit if it approaches a specific number

A function has an infinite limit if it increases or decreases without bound as x approaches a value.

A function has an infinite limit if it equals infinity at some point

A function has an infinite limit if it oscillates between two values

Infinite limits describe the behavior of a function as x approaches a particular value where the function increases or decreases without bound. The function does not approach a specific finite number but rather grows arbitrarily large (positive or negative infinity) as x gets closer to the target value.

Explanation

Infinite limits describe the behavior of a function as x approaches a particular value where the function increases or decreases without bound. The function does not approach a specific finite number but rather grows arbitrarily large (positive or negative infinity) as x gets closer to the target value.

8) Evaluate limx→-1⁺ 2/(x+1)

0

2

+∞

-∞

As x approaches -1 from the right (x→-1⁺), the denominator (x+1) approaches 0 from the positive side. When we have 2 divided by a very small positive number, the result becomes very large and positive.

Explanation

As x approaches -1 from the right (x→-1⁺), the denominator (x+1) approaches 0 from the positive side. When we have 2 divided by a very small positive number, the result becomes very large and positive.

9) For f(x) = 1/(x-3), what is lim_{x→3⁻} f(x)?

0

1/3

+∞

-∞

As x approaches 3 from the left (x→3⁻), x is less than 3, so (x-3) is negative and approaches 0. When we divide 1 by a very small negative number, the result becomes very large and negative.

Explanation

As x approaches 3 from the left (x→3⁻), x is less than 3, so (x-3) is negative and approaches 0. When we divide 1 by a very small negative number, the result becomes very large and negative.

10) Which function demonstrates vertical asymptote behavior at x = 0?

F(x) = x/(x+1)

F(x) = x²/(x+1)

F(x) = 1/x

F(x) = (x-1)/(x+1)

For f(x) = 1/x, when x approaches 0, the denominator approaches 0 while the numerator remains 1. This causes the function values to become arbitrarily large in magnitude, creating a vertical asymptote at x = 0.

Explanation

For f(x) = 1/x, when x approaches 0, the denominator approaches 0 while the numerator remains 1. This causes the function values to become arbitrarily large in magnitude, creating a vertical asymptote at x = 0.

11) Evaluate lim_{x→5⁺} 3/(5-x)

0

3/5

+∞

-∞

As x approaches 5 from the right (x→5⁺), the expression (5-x) approaches 0 from the negative side because x is greater than 5, making (5-x) negative. When we divide 3 by a very small negative number, the result becomes very large and negative.

Explanation

As x approaches 5 from the right (x→5⁺), the expression (5-x) approaches 0 from the negative side because x is greater than 5, making (5-x) negative. When we divide 3 by a very small negative number, the result becomes very large and negative.

12) What does it mean for a function to have a vertical asymptote at x = a?

The function equals a at x = a

The function is undefined at x = a but approaches ±∞ as x approaches a

The function has a hole at x = a

The function equals infinity at x = a

A vertical asymptote at x = a means the function is not defined at x = a (typically because of division by zero), and as x gets closer to a, the function values increase or decrease without bound, approaching positive or negative infinity.

Explanation

A vertical asymptote at x = a means the function is not defined at x = a (typically because of division by zero), and as x gets closer to a, the function values increase or decrease without bound, approaching positive or negative infinity.

13) Evaluate lim_{x→2⁻} 1/(2-x)

0

1/2

+∞

-∞

As x approaches 2 from the left (x→2⁻), the expression (2-x) approaches 0 from the positive side because x is less than 2, making (2-x) positive. When we divide 1 by a very small positive number, the result becomes very large and positive.

Explanation

As x approaches 2 from the left (x→2⁻), the expression (2-x) approaches 0 from the positive side because x is less than 2, making (2-x) positive. When we divide 1 by a very small positive number, the result becomes very large and positive.

14) Which of the following values of x is a vertical asymptote for the function f(x) = 1/(x²-9)?

X = 0

X = 3

X = 9

X = -9

A vertical asymptote occurs where the denominator equals zero and the numerator is not zero. For f(x) = 1/(x²-9), the denominator equals zero when x²-9 = 0, which gives x = 3 or x = -3. Both values create vertical asymptotes.

Explanation

A vertical asymptote occurs where the denominator equals zero and the numerator is not zero. For f(x) = 1/(x²-9), the denominator equals zero when x²-9 = 0, which gives x = 3 or x = -3. Both values create vertical asymptotes.

15) Evaluate lim_{x→1⁺} 1/(x-1)²

0

1

+∞

-∞

As x approaches 1 from the right (x→1⁺), the denominator (x-1)² approaches 0 from the positive side (since squaring always gives a positive result). When we divide 1 by a very small positive number, the result becomes very large and positive.

Explanation

As x approaches 1 from the right (x→1⁺), the denominator (x-1)² approaches 0 from the positive side (since squaring always gives a positive result). When we divide 1 by a very small positive number, the result becomes very large and positive.

Infinite Limits and Vertical Asymptotes