Limits: l'Hopital's rule; Maclaurin practice questions.

Determine lim (x→0) of (sin x − x)/x³.

• 1/6
• 0
• −1/6
• −1/3

Answer: −1/6

Using the Maclaurin series sin x = x − x³/6 + …, the numerator is −x³/6 + higher-order terms, so the limit is −1/6.

Determine lim (x→0) of (eˣ − 1 − x)/x².

• 1/2
• 0
• 1
• The limit does not exist

Answer: 1/2

Since eˣ = 1 + x + x²/2 + …, the numerator equals x²/2 + higher-order terms, giving a limit of 1/2. Two applications of l'Hôpital's rule give the same result: eˣ/2 → 1/2.

State the first three non-zero terms of the Maclaurin series for ln(1 + x).

• x + x²/2 + x³/3
• 1 + x − x²/2
• x − x² + x³
• x − x²/2 + x³/3

Answer: x − x²/2 + x³/3

Differentiating repeatedly or integrating the geometric series 1/(1+x) = 1 − x + x² − … term by term gives ln(1+x) = x − x²/2 + x³/3 − …. The signs alternate — a common error is to keep them all positive.

State the coefficient of x⁴ in the Maclaurin series for cos x.

• −1/24
• 1/24
• 1/4
• −1/4

Answer: 1/24

cos x = 1 − x²/2! + x⁴/4! − …, so the coefficient of x⁴ is +1/4! = 1/24. The signs alternate starting with +1, so the x⁴ term is positive.

Determine lim (x→∞) of x·e⁻ˣ.

• 1
• ∞
• 0
• e

Answer: 0

Rewrite as x/eˣ, an ∞/∞ form; l'Hôpital's rule gives 1/eˣ → 0. Exponential growth dominates polynomial growth.

To which of the following indeterminate forms can l'Hôpital's rule be applied directly, without first rewriting the expression?

• 0/0
• 0 × ∞
• ∞ − ∞
• 1^∞

Answer: 0/0

L'Hôpital's rule applies directly only to quotients of the form 0/0 or ∞/∞. Forms such as 0 × ∞, ∞ − ∞ and 1^∞ must first be rewritten as a quotient (often using logarithms).