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

$\binom{5}{3} = \dfrac{5!}{3! \cdot 2!} = 10$.

$\bullet$ Rewrite as an exponential:

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

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

$$\lim_{x\to\infty}\frac{\ln x}{\sqrt{x}}=\lim_{x\to\infty}\frac{2}{\sqrt{x}}=0$$

$\bullet$ Exponentiate:

$$L=e^0=1$$

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

$f$ 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 $2\sin 20°$ and use $2\sin\theta\cos\theta=\sin 2\theta$:

$$\cos 20°\cos 40°\cos 80° = \dfrac{\sin 40°\cos 40°\cos 80°}{2\sin 20°}$$

$\bullet$ Apply the identity again:

$$= \dfrac{\sin 80°\cos 80°}{4\sin 20°}$$

$\bullet$ And once more:

$$= \dfrac{\sin 160°}{8\sin 20°}$$

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

$$= \dfrac{\sin 20°}{8\sin 20°} = \dfrac{1}{8}$$

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

Euler's classical result: $\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{n^2} = \dfrac{\pi^2}{6}$.

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

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 $\theta : 0 \to \pi/4$:

$$I = \int_0^{\pi/4} \ln(1 + \tan\theta)\,d\theta.$$

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

$$1 + \tan\theta \;=\; 1 + \frac{1 - \tan\phi}{1 + \tan\phi} \;=\; \frac{2}{1 + \tan\phi}.$$

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

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

$$I \;=\; \int_0^{\pi/4} \bigl[\ln 2 - \ln(1 + \tan\theta)\bigr]\,d\theta \;=\; \frac{\pi}{4}\ln 2 \;-\; I.$$

Solve. $\;2I = \dfrac{\pi}{4}\ln 2$, so

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

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