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

$\bullet$ 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:

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

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:

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

$\bullet$ Apply the Pythagorean identity:

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

$\bullet$ Taking the negative root:

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

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

Middle coefficient: $\left(\genfrac{}{}{0px}{}{8}{4}\right)=70$.

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

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

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

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

$P=\left(\genfrac{}{}{0px}{}{4}{2}\right)\text{\hspace{0.17em}⁣}{\left({\displaystyle \frac{1}{3}}\right)}^2\text{\hspace{0.17em}⁣}{\left({\displaystyle \frac{2}{3}}\right)}^2=6\cdot {\displaystyle \frac{1}{9}}\cdot {\displaystyle \frac{4}{9}}={\displaystyle \frac{24}{81}}={\displaystyle \frac{8}{27}}$.