Dokažte nerovnosti
\( | \sin x − \sin y| \leq |x − y| \) abs(sinx - siny) <= abs(x - y) RegionPlot(abs(sin(x)-sin(y)), abs(x-y))
\( e^x > 1 + x pro x \ne 0 \) e^x > 1+x
\( x - \frac{x^2}{2} < \ln (1 + x) < x pro x \gt 0 \)
\( x - \frac{x^3}{6} < \sin x < x pro x \gt 0 \) x - x^3/6 < sin x < x from x<0