Daily math problems for August 14, 2026 — Limits, Logaritmi, Trigonometry & more

$\bullet$ Use the fundamental limits:

$$\lim_{u\to 0}\dfrac{e^u-1}{u}=1,\qquad \lim_{u\to 0}\dfrac{\sin u}{u}=1$$

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

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

$\bullet$ Take the limit:

$$1\cdot 1\cdot\dfrac{2}{3}=\dfrac{2}{3}$$

Avem $\log_5 5^{12} = 12$ și $\log_7 1 = 0$. Deci expresia este $12 - 0 = 12$.

$\bullet$ Compute the determinant directly:

$$\cos^2 x-\sin^2 x=\cos 2x$$

$\bullet$ From the hypothesis:

$$\cos 2x=\tfrac{3}{5}$$

$\bullet$ Apply the Pythagorean identity:

$$\sin^2 2x=1-\tfrac{9}{25}=\tfrac{16}{25}$$

$\bullet$ Taking the negative root:

$$\sin 2x=-\tfrac{4}{5}$$

$\sin^2\theta = 1 - \dfrac{4}{9} = \dfrac{5}{9}$, so $\sin\theta = \dfrac{\sqrt{5}}{3}$. Hence $\tan\theta = \dfrac{\sin\theta}{\cos\theta} = \dfrac{\sqrt{5}/3}{2/3} = \dfrac{\sqrt{5}}{2}$.

Middle coefficient: $\binom{8}{4} = 70$.

Cauchy's equation with continuity gives $f(x) = cx$ for some constant $c$. Since $f(1) = 3$, $c = 3$, so $f(5) = 15$.

$S_{12} = \dfrac{12\,(2 \cdot 50 + 11 \cdot 10)}{2} = 6 \cdot 210 = 1260$.

For any $x \in [-1, 1]$, $\arcsin x + \arccos x = \dfrac{\pi}{2}$ (a standard identity, since $\arccos x = \dfrac{\pi}{2} - \arcsin x$).

$e^{a + b} = e^a e^b$ ✓ (homomorphism). $e^x$ is strictly increasing (injective) and surjective onto $(0, \infty)$. Hence an isomorphism. (Inverse: $\ln$.)

$P = \binom{4}{2}\!\left(\dfrac{1}{3}\right)^2 \!\left(\dfrac{2}{3}\right)^2 = 6 \cdot \dfrac{1}{9} \cdot \dfrac{4}{9} = \dfrac{24}{81} = \dfrac{8}{27}$.