Задача №2029

\[ \lim_{x\to{0}}\frac{3\tg{4x}-12\tg{x}}{3\sin{4x}-12\sin{x}} =\left[\frac{0}{0}\right] =\lim_{x\to{0}}\frac{\left(3\tg{4x}-12\tg{x}\right)'}{\left(3\sin{4x}-12\sin{x}\right)'} =\lim_{x\to{0}}\frac{\frac{1}{\cos^2{4x}}-\frac{1}{\cos^2{x}}}{\cos{4x}-\cos{x}}=\\ =\lim_{x\to{0}}\frac{-\left(\cos{4x}-\cos{x}\right)\cdot\left(\cos{x}+\cos{4x}\right)}{\cos^2{x}\cos^2{4x}\cdot\left(\cos{4x}-\cos{x}\right)} =\lim_{x\to{0}}\frac{-\left(\cos{x}+\cos{4x}\right)}{\cos^2{x}\cos^2{4x}} =-2. \]