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Limits of Sequences

12 practice questions with full worked solutions. Free, no account needed.

Problems & worked solutions

Problem #0523 International

Problem 1 — Limits of Sequences

The limit lim⁡n→∞2n\displaystyle\lim_{n \to \infty} 2^n equals:

Show answer & worked solution
  1. A. 00
  2. B. 11
  3. C. ∞\infty✓ correct
  4. D. does not exist

2n2^n grows without bound: lim⁡2n=∞\lim 2^n = \infty.

Problem #0522 International

Problem 2 — Limits of Sequences

The limit lim⁡n→∞ ⁣(12)n\displaystyle\lim_{n \to \infty} \!\left(\dfrac{1}{2}\right)^n equals:

Show answer & worked solution
  1. A. 00✓ correct
  2. B. 12\dfrac{1}{2}
  3. C. 11
  4. D. ∞\infty

 ⁣(12)n→0\!\left(\dfrac{1}{2}\right)^n \to 0 since ∣q∣=12<1|q| = \dfrac{1}{2} < 1.

Problem #0521 International

Problem 3 — Limits of Sequences

The limit lim⁡n→∞1n\displaystyle\lim_{n \to \infty} \dfrac{1}{n} equals:

Show answer & worked solution
  1. A. 00✓ correct
  2. B. 11
  3. C. ∞\infty
  4. D. does not exist

As n→∞n \to \infty, 1n→0\dfrac{1}{n} \to 0.

Problem #0763 FR Spé

Problem 4 — Limits of Sequences

A sequence (un)(u_n) is defined by u0=1u_0 = 1 and, for every integer n≥0n \ge 0, un+1=12un+3u_{n+1} = \dfrac{1}{2}u_n + 3. Given that (un)(u_n) converges, its limit equals:

Show answer & worked solution
  1. A. 33
  2. B. 66✓ correct
  3. C. 32\dfrac{3}{2}
  4. D. −6-6

Since (un)(u_n) converges to some limit ℓ\ell, taking the limit on both sides of the recurrence gives the fixed-point equation: ℓ=12ℓ+3\ell = \frac{1}{2}\ell + 3 12ℓ=3\frac{1}{2}\ell = 3 ℓ=6\ell = 6

Problem #0527 International

Problem 5 — Limits of Sequences

The limit lim⁡n→∞ ⁣(1+1n)n\displaystyle\lim_{n \to \infty} \!\left(1 + \dfrac{1}{n}\right)^n equals:

Show answer & worked solution
  1. A. 00
  2. B. 11
  3. C. ee✓ correct
  4. D. ∞\infty

This is the standard definition of ee:  ⁣(1+1n)n→e\!\left(1 + \dfrac{1}{n}\right)^n \to e.

Problem #0528 International

Problem 6 — Limits of Sequences

The limit lim⁡n→∞n3+2nn4+1\displaystyle\lim_{n \to \infty} \dfrac{n^3 + 2n}{n^4 + 1} equals:

Show answer & worked solution
  1. A. 00✓ correct
  2. B. 11
  3. C. 22
  4. D. ∞\infty

The denominator has higher degree, so the ratio tends to 00.

Problem #0525 International

Problem 7 — Limits of Sequences

The limit lim⁡n→∞n+1n2+3\displaystyle\lim_{n \to \infty} \dfrac{n + 1}{n^2 + 3} equals:

Show answer & worked solution
  1. A. 00✓ correct
  2. B. 11
  3. C. 13\dfrac{1}{3}
  4. D. ∞\infty

n+1n2+3=1/n+1/n21+3/n2→01=0\dfrac{n + 1}{n^2 + 3} = \dfrac{1/n + 1/n^2}{1 + 3/n^2} \to \dfrac{0}{1} = 0.

Problem #0524 International

Problem 8 — Limits of Sequences

The limit lim⁡n→∞3n2+52n2−1\displaystyle\lim_{n \to \infty} \dfrac{3n^2 + 5}{2n^2 - 1} equals:

Show answer & worked solution
  1. A. 00
  2. B. 11
  3. C. 32\dfrac{3}{2}✓ correct
  4. D. ∞\infty

3n2+52n2−1=3+5/n22−1/n2→32\dfrac{3n^2 + 5}{2n^2 - 1} = \dfrac{3 + 5/n^2}{2 - 1/n^2} \to \dfrac{3}{2}.

Problem #0526 International

Problem 9 — Limits of Sequences

The limit lim⁡n→∞sin⁡nn\displaystyle\lim_{n \to \infty} \dfrac{\sin n}{n} equals:

Show answer & worked solution
  1. A. 00✓ correct
  2. B. 11
  3. C. 1n\dfrac{1}{n}
  4. D. does not exist

By the squeeze theorem, −1n≤sin⁡nn≤1n-\dfrac{1}{n} \le \dfrac{\sin n}{n} \le \dfrac{1}{n} and both bounds tend to 00, so the limit is 00.

Problem #0530 International

Problem 10 — Limits of Sequences

The limit lim⁡n→∞ ⁣(1+2n)n\displaystyle\lim_{n \to \infty} \!\left(1 + \dfrac{2}{n}\right)^n equals:

Show answer & worked solution
  1. A. 11
  2. B. ee
  3. C. e2e^2✓ correct
  4. D. ∞\infty

With a=2a = 2:  ⁣(1+2n)n→e2\!\left(1 + \dfrac{2}{n}\right)^n \to e^2.

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