AP Calculus AB Infinite Limits and Asymptotes Quiz: 10 Questions with Explanations

A \(-\infty\)B \(\infty\)C \(0\)D \(1\)

Just left of 1, x – 1 is a small negative number, so the quotient is a large negative number.

A \(-\infty\)B \(0\)C does not exist and is not infiniteD \(\infty\)

The square is positive from both sides, so both one-sided limits are +infinity.

A \(y=-2\)B \(y=-\frac{1}{2}\)C \(y=4\)D \(x=4\)

Equal degrees give the ratio of leading coefficients, 4/1. The line x = -2 is the vertical asymptote, not horizontal.

A \(0\)B \(\infty\)C \(-\frac{2}{7}\)D \(5\)

The degrees are equal, so the limit is 5/1. The other choices ignore the leading terms.

A \(2\)B \(\infty\)C \(0\)D \(-9\)

The denominator has the higher degree, so it wins and the limit is 0.

A \(\frac{x}{x-3}\)B \(\frac{x-3}{x-3}\)C \(\frac{x^{2}-9}{x-3}\)D \(\frac{3}{x+3}\)

Only x/(x – 3) has a nonzero numerator at x = 3. The other two simplify and leave a hole or no break at all.

A \(0\)B \(2\)C \(5\)D \(\infty\)

The exponential term approaches 0, leaving 2. The value 5 is f(0).

A \(0\)B \(\infty\)C \(1\)D \(-\infty\)

As x becomes very negative, e^x is a tiny positive number, so the limit is 0.

A \(1\)B \(0\)C \(\infty\)D does not exist because cosine oscillates

Cosine is bounded between -1 and 1 while x grows, so the Squeeze Theorem gives 0.

A \(4\)B \(-4\)C \(-2\)D \(\infty\)

The degrees are equal, so the limit is the ratio of leading coefficients, 4/(-1) = -4. Ignoring the negative sign in the denominator gives 4.