Daily math problems for August 5, 2026 — Combinatorics, Limits, Derivatives & more

$\left(\genfrac{}{}{0px}{}{5}{3}\right)={\displaystyle \frac{5!}{3!\cdot 2!}}=10$.

$\bullet$ Rewrite as an exponential:

$$x^{1\mathrm{/}\sqrt{x}}=e^{\ln x\mathrm{/}\sqrt{x}}$$

$\bullet$ Apply L'Hôpital to the exponent:

$$\underset{x\to \mathrm{\infty }}{\lim }\frac{\ln x}{\sqrt{x}}=\underset{x\to \mathrm{\infty }}{\lim }\frac{2}{\sqrt{x}}=0$$

$\bullet$ Exponentiate:

$$L=e^0=1$$

$f^{\mathrm{\prime }}(x)=2e^{2x}$, so $f^{\mathrm{\prime }}(0)=2\cdot e^0=2$.

$f$ is strictly increasing on $\mathbb{R}$, hence injective. For every $y\in \mathbb{R}$, $x=\sqrt[3]{y}$ satisfies $f(x)=y$, hence surjective. So $f$ is bijective, with $f^{-1}(y)=\sqrt[3]{y}$.

Set $\sqrt{x^2+y^2}=\sqrt{(x-4{)}^2+y^2}$. Squaring: $x^2=(x-4{)}^2=x^2-8x+16$, so $8x=16$, $x=2$.

$\bullet$ Multiply numerator and denominator by $2\sin 20\mathrm{°}$ and use $2\sin \theta \cos \theta =\sin 2\theta$:

$$\cos 20\mathrm{°}\cos 40\mathrm{°}\cos 80\mathrm{°}={\displaystyle \frac{\sin 40\mathrm{°}\cos 40\mathrm{°}\cos 80\mathrm{°}}{2\sin 20\mathrm{°}}}$$

$\bullet$ Apply the identity again:

$$={\displaystyle \frac{\sin 80\mathrm{°}\cos 80\mathrm{°}}{4\sin 20\mathrm{°}}}$$

$\bullet$ And once more:

$$={\displaystyle \frac{\sin 160\mathrm{°}}{8\sin 20\mathrm{°}}}$$

$\bullet$ Use $\sin 160\mathrm{°}=\sin (180\mathrm{°}-20\mathrm{°})=\sin 20\mathrm{°}$:

$$={\displaystyle \frac{\sin 20\mathrm{°}}{8\sin 20\mathrm{°}}}={\displaystyle \frac{1}{8}}$$

$\overrightarrow{AB}=(4,0)$ and $\overrightarrow{DC}=(4,0)$ — same vector, so $ABCD$ is a parallelogram. Base $=4$, height $=3$. Area $=12$.

Euler's classical result: ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}{\displaystyle \frac{1}{n^2}}={\displaystyle \frac{{\pi }^2}{6}}}$.

Direct substitution gives $\frac{0}{0}$. Differentiate top and bottom: ${\displaystyle \underset{x\to 0}{\lim }{\displaystyle \frac{e^x-1}{2x}}}$ — still $\frac{0}{0}$. Once more: ${\displaystyle \underset{x\to 0}{\lim }{\displaystyle \frac{e^x}{2}}={\displaystyle \frac{1}{2}}}$.

Substitute $x=\tan \theta$. Then $dx={\sec }^2\theta d\theta$ and $1+x^2={\sec }^2\theta$, so the ${\sec }^2\theta$ on top and bottom cancel cleanly. The bounds $0\to 1$ map to $\theta :0\to \pi \mathrm{/}4$:

$$I={\int }_0^{\pi \mathrm{/}4}\ln (1+\tan \theta )d\theta \mathrm{.}$$

Reflect by $\varphi =\pi \mathrm{/}4-\theta$. The angle-difference formula gives $\tan (\pi \mathrm{/}4-\varphi )={\displaystyle \frac{1-\tan \varphi }{1+\tan \varphi }}$, so

$$1+\tan \theta =1+\frac{1-\tan \varphi }{1+\tan \varphi }=\frac{2}{1+\tan \varphi }\mathrm{.}$$

Taking logs: $\ln (1+\tan \theta )=\ln 2-\ln (1+\tan \varphi )$.

Pair the integral with its reflected twin. Renaming the dummy variable $\varphi \to \theta$ (the bounds are unchanged because the substitution is a reflection of $[0,\pi \mathrm{/}4]$ onto itself):

$$I={\int }_0^{\pi \mathrm{/}4}[\ln 2-\ln (1+\tan \theta )]d\theta =\frac{\pi }{4}\ln 2-I\mathrm{.}$$

Solve. $2I={\displaystyle \frac{\pi }{4}}\ln 2$, so

$$\boxed{{\displaystyle I=\frac{\pi }{8}\ln 2\mathrm{.}}}$$

The reflection trick — pairing $f(\theta )$ with $f(\pi \mathrm{/}4-\theta )$ — works whenever the integrand simplifies under that reflection. Worth keeping in your toolbox alongside the half-angle and Weierstrass substitutions.