Question

РЕШИТЕ ПРЕДЕЛЫ ПРАВИЛОМ ЛОПИТАЛЯ

Answer 1

$\displaystyle 1)~\lim_{x\to 0}\dfrac{\ln\cos 2x}{\sin^2 5x}=\bigg[\dfrac{0}{0}\bigg]\overset{\text{L'H}}{=}\lim_{x\to 0}\dfrac{\frac{-2\sin 2x}{\cos 2x}}{2\cdot\sin 5x\cdot\cos 5x\cdot 5}=\dfrac{-1\cdot\frac{2}{5}}{5}=-\dfrac{2}{25}$

$\displaystyle 2)~\lim_{x\to 0}\dfrac{e^{2x}-2x-1}{\sin^2 3x}=\bigg[\dfrac{0}{0}\bigg]\overset{\text{L'H}}{=}\lim_{x\to 0}\dfrac{2e^{2x}-2}{2\cdot\sin 3x\cdot \cos 3x\cdot 3}=\medskip \\=\lim_{x\to 0}\dfrac{e^{2x}-1}{3\cdot\sin 3x\cos 3x}=\lim_{x\to 0}\dfrac{1+2x+o(x)-1}{3\cdot\sin 3x\cos 3x}=\dfrac{2}{9}$