Let A = {1, 2, 3}, B = {2, 3, 4}, with universal set U = {1, 2, 3, 4, 5}. Compute A ∩ B.

A ∩ B = {2, 3}. The complement in U = {1, 2, 3, 4, 5} is {1, 4, 5}.

lim_n → ∞ ((n!)^(1/n))/(n) equals:

By Stirling, (n!)^(1/n) ~ (n)/(e) (2π n)^1/(2n). The factor (2π n)^1/(2n) → 1, so ((n!)^(1/n))/(n) → (1)/(e).

Daily Math Problems — May 23, 2026: Sets, Trigonometry, Limits

dailymath

Problems & worked solutions

RO M1

Problem 1 — Sets of Real Numbers

Let $A = {1, 2, 3}$ , $B = {2, 3, 4}$ , with universal set $U = {1, 2, 3, 4, 5}$ . Compute $\overline{A \cap B}$ .

Show answer & worked solution

A. ${2, 3}$
B. ${1, 4, 5}$ ✓ correct
C. ${1, 2, 4}$
D. ${1, 2, 3, 4, 5}$

$A \cap B = {2, 3}$ . The complement in $U = {1, 2, 3, 4, 5}$ is ${1, 4, 5}$ .

RO M1

Problem 2 — Trigonometric Equations

How many solutions does $2 \cos^{2} x = 1$ have on $[0, 2 π)$ ?

Show answer & worked solution

A. $2$
B. $3$
C. $4$ ✓ correct
D. $6$

$\cos^{2} x = \frac{1}{2} \Rightarrow \cos x = \pm \frac{\sqrt{2}}{2}$ . Solutions: $\frac{π}{4}, \frac{3 π}{4}, \frac{5 π}{4}, \frac{7 π}{4}$ .

International

Problem 3 — Calculus

$\lim_{n \to \infty} \frac{\sqrt[n]{n!}}{n}$ equals:

Show answer & worked solution

A. $\frac{1}{e}$ ✓ correct
B. $1$
C. $e$
D. $0$

By Stirling, $\sqrt[n]{n!} \sim \frac{n}{e} (2 π n)^{1 / (2 n)}$ . The factor $(2 π n)^{1 / (2 n)} \to 1$ , so $\frac{\sqrt[n]{n!}}{n} \to \frac{1}{e}$ .

Problems & worked solutions

RO M1

Problem 1 — Sets of Real Numbers

Let $A = {1, 2, 3}$ , $B = {2, 3, 4}$ , with universal set $U = {1, 2, 3, 4, 5}$ . Compute $\overline{A \cap B}$ .

Show answer & worked solution

A. ${2, 3}$
B. ${1, 4, 5}$ ✓ correct
C. ${1, 2, 4}$
D. ${1, 2, 3, 4, 5}$

$A \cap B = {2, 3}$ . The complement in $U = {1, 2, 3, 4, 5}$ is ${1, 4, 5}$ .

RO M1

Problem 2 — Trigonometric Equations

How many solutions does $2 \cos^{2} x = 1$ have on $[0, 2 π)$ ?

Show answer & worked solution

A. $2$
B. $3$
C. $4$ ✓ correct
D. $6$

$\cos^{2} x = \frac{1}{2} \Rightarrow \cos x = \pm \frac{\sqrt{2}}{2}$ . Solutions: $\frac{π}{4}, \frac{3 π}{4}, \frac{5 π}{4}, \frac{7 π}{4}$ .

International

Problem 3 — Calculus

$\lim_{n \to \infty} \frac{\sqrt[n]{n!}}{n}$ equals:

Show answer & worked solution

A. $\frac{1}{e}$ ✓ correct
B. $1$
C. $e$
D. $0$

By Stirling, $\sqrt[n]{n!} \sim \frac{n}{e} (2 π n)^{1 / (2 n)}$ . The factor $(2 π n)^{1 / (2 n)} \to 1$ , so $\frac{\sqrt[n]{n!}}{n} \to \frac{1}{e}$ .