Question

URGENT CALCULATI LIMITA ASTA CU L'HOPITAL !!!! va rog mult am test maine

Answer 1

Răspuns:

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Answer 2

$\underset{x>0}{\lim\limits_{x\to 0}}\,\dfrac{\ln(\sin 2x)}{\ln(\sin 3x)}\overset{\frac{-\infty}{-\infty}}{=} \underset{x>0}{\lim\limits_{x\to 0}}\,\dfrac{\dfrac{(\sin2x)'}{\sin 2x}}{\dfrac{(\sin3x)'}{\sin 3x}} =\underset{x>0}{\lim\limits_{x\to 0}}\,\dfrac{\dfrac{2\cos(2x)}{\sin 2x}}{\dfrac{3\cos(3x)}{\sin 3x}}=\\ \\ \\= \underset{x>0}{\lim\limits_{x\to 0}}\,\Big(\dfrac{2\cos2x}{3\cos 3x}\cdot \dfrac{\sin 3x}{\sin 2x}\Big)=$

$=\underset{x>0}{\lim\limits_{x\to 0}}\,\left[\dfrac{2\cos2x}{3\cos 3x}\cdot \Big(\dfrac{\sin 3x}{3x}\Big)\cdot \Big(\dfrac{2x}{\sin 2x}\Big)\cdot \dfrac{3}{2}\right] = \\ \\ \\=\dfrac{2\cdot 1}{3\cdot 1}\cdot 1\cdot 1\cdot \dfrac{3}{2} = \boxed{1}$

$\lim\limits_{u\to 0}\dfrac{\sin u}{u} = \lim\limits_{u\to 0}\dfrac{u}{\sin u}=1$