Limit assumed to exist

Reasoning with a limit before establishing there is one. limsup is not lim.

Would you have fallen for these?

Each is a real option from a previous-year paper. Read it cold and decide before opening it.

• Option A

  1

  Why this is wrong, and why it looks right

  For odd $n, 2+(-1)^{n}=1$, so $a_{n}=1$ and |$a_{n}$|^(1/n)=1 along that subsequence — tempting to call this 'the' limit and get R=1/1=1. But even-n terms give $a_{n}=3^{n}$, so |$a_{n}$|^(1/n)=3 infinitely often; Cauchy–Hadamard uses the limsup over all n, which is 3, not the value along a single subsequence.

  From this question

  The radius of convergence of $\sum_{n\ge1}$ $(2 + (-1)^{n})^{n} x^{n}$ is

  Part B · free

• Option C

  1/2

  Why this is wrong, and why it looks right

  This treats $2+(-1)^{n}$ as if it settled down to its average value 2 (since (-1)^{n} 'averages' to 0), giving $a_{n}\approx2^{n}$ and R=1/2. But $2+(-1)^{n}$ doesn't converge at all — it alternates exactly between 1 and 3 forever, and the root test needs the limsup of the true extreme value 3, not a fictitious average.

  From this question

  The radius of convergence of $\sum_{n\ge1}$ $(2 + (-1)^{n})^{n} x^{n}$ is

  Part B · free

• Option D

  {$a_{n}$} converges.

  Why this is wrong, and why it looks right

  Convergence requires $\limsup a_{n} = \liminf a_{n}$, but here 1 ≠ −1: the odd and even terms approach different values, so the sequence oscillates forever instead of settling on one number.

  From this question

  Let $a_{n} = (-1)^{n}(1 + 1/n)$ for n ≥ 1. Which of the following are true?

  Part C · free

• Option C

  1 and 0

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B

• Option D

  0 and −1

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B

• Option C

  0

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B

• Option C

  become neutral.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part A · december 2024

• Option A

  $\lim \inf(n\to\infty) x_{n} \ge 2025$

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2024

• Option B

  $\lim \sup(n\to\infty) x_{n} \le 2025$

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2024

• Option D

  $\lim \inf(k\to\infty) x_(n_{k}) > 2025$

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2024

• Option C

  3/4

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2024

• Option A

  A = 0

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2024

• Option D

  $\lim \sup(n\to\infty) 1/(\sqrt{n + \sqrt{n}} - \sqrt{n}) = 3$

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option D

  converges to 0

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B

• Option B

  1/2

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2024

• Option B

  $N_{n}$ converges to e⁻$^{2}$ in probability.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option B

  (e, e)

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B

• Option B

  $lim_{x\to\infty}$ (x + sin x)/x

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C

• Option D

  f extends continuously to [0, 1]

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C

• Option C

  For every $x \in \mathbb{R}$ there exists some n such that $b_{n} > x$

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2024

• Option D

  For every $x \in \mathbb{R}$ there exists some n such that $b_{n} < x$

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2024

• Option D

  $P^{n}$ converges entrywise as $n \to \infty$.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C

• Option C

  $lim_{x\to\infty}$ x f′(x) need not exist.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2025

• Option A

  $\lim(n\to\infty)$ |$X_{i}(n)$|/n exists for all 1 ≤ i ≤ 8.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option D

  $\lim \sup(n\to\infty)$ |$X_{i}(n)$|/n < 1/8 for all 1 ≤ i ≤ 8.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option B

  $\lim \sup(n\to\infty) a_{n} = \lim(n\to\infty) b_{n}$

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option C

  $\lim \inf(n\to\infty) a_{n} = \lim(n\to\infty) c_{n}$

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option D

  $\lim(n\to\infty) b_{n} = \lim(n\to\infty) c_{n}$

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option A

  If $lim_{n}\to\infty b_{n}$ exists in $\mathbb{R}$, then {$a_{n} : n \ge 1$} is bounded

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2024

• Option B

  If $lim_{n}\to\infty b_{n} = 1$, then $lim_{n}\to\infty a_{n}$ exists in $\mathbb{R}$

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2024

• Option B

  if inf{$a_{n}$ | n ≥ 1} $= limsup_{n}\to\infty a_{n}$, then $(a_{n})$ is convergent

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · june 2024

• Option A

  $(a_{n})_{n}\ge_{1}$ converges but $(b_{n})_{n}\ge_{1}$ does not converge.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2024

• Option B

  $(b_{n})_{n}\ge_{1}$ converges but $(a_{n})_{n}\ge_{1}$ does not converge.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2024

• Option D

  Neither $(a_{n})_{n}\ge_{1}$ nor $(b_{n})_{n}\ge_{1}$ converges.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2024

• Option A

  $\limsup a_{n} = \sqrt{3}/2$.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2023

• Option C

  $Y_{2}$ follows beta distribution with $E(Y_{2}) = 1/2$.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option D

  $f_{n} \to 0$ uniformly and $f_{n}' \to 0$ pointwise but not uniformly

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B

The counterexamples that settle these

One example is enough to kill a false statement. These are the standard ones for the claims above.

• Continuous on a bounded interval ⇒ bounded
• Uniformly continuous ⇒ Lipschitz
• Differentiable ⇒ continuously differentiable
• Pointwise limit of continuous functions is continuous
• Σaₙ convergent ⇒ Σaₙ² convergent
• aₙ → 0 ⇒ Σaₙ converges

Other mistakes that cost marks