Daily math problems for September 23, 2026 — Calculus, Limits, Algebra & more

$F(x)={\displaystyle \frac{x^3}{3}}$. $F(2)-F(1)={\displaystyle \frac{8}{3}}-{\displaystyle \frac{1}{3}}={\displaystyle \frac{7}{3}}$.

$\bullet$ Use the fundamental limits:

$$\underset{u\to 0}{\lim }{\displaystyle \frac{e^u-1}{u}}=1,\underset{u\to 0}{\lim }{\displaystyle \frac{\sin u}{u}}=1$$

$\bullet$ Force the arguments to match the denominators:

$${\displaystyle \frac{e^{2x}-1}{\sin 3x}}={\displaystyle \frac{e^{2x}-1}{2x}}\cdot {\displaystyle \frac{3x}{\sin 3x}}\cdot {\displaystyle \frac{2x}{3x}}$$

$\bullet$ Take the limit:

$$1\cdot 1\cdot {\displaystyle \frac{2}{3}}={\displaystyle \frac{2}{3}}$$

A semigroup is a set with an associative binary operation.

Associative ✓, identity $1$ ✓. But $2$ has no integer multiplicative inverse. Hence a monoid but not a group.

$\sin {\displaystyle \frac{\pi }{3}}={\displaystyle \frac{\sqrt{3}}{2}}$ and $\cos {\displaystyle \frac{\pi }{6}}={\displaystyle \frac{\sqrt{3}}{2}}$. Sum: $\sqrt{3}$.

$3x+4y=25$.

$\cos A={\displaystyle \frac{25+9-49}{2\cdot 5\cdot 3}}={\displaystyle \frac{-15}{30}}=-{\displaystyle \frac{1}{2}}$. Since $A\in (0,\pi )$, $A={\displaystyle \frac{2\pi }{3}}$.

$2x^2+x+1=2(x^2-x+2)\Rightarrow 2x^2+x+1=2x^2-2x+4\Rightarrow 3x=3\Rightarrow x=1$.

Verify the domain: at $x=1$, $2x^2+x+1=4>0$ and $x^2-x+2=2>0$, so both logs are defined. The solution set is $\{1\}$.

$y^2-6y+8=0\Rightarrow y\in \{2,4\}$. Then $x^2=2\Rightarrow x=\pm \sqrt{2}$ and $x^2=4\Rightarrow x=\pm 2$. Four distinct real solutions.

Since $b^2=ac$, the product $abc=b\cdot ac=b\cdot b^2=b^3=64$, so $b=4$.