LimitesAche o limite, na figura.... Pergunta de ideia deMinhascontasnet

jbsenajr

Resposta:

Explicação passo-a-passo:

Usando a Regra de L'Hopital

$\lim_{x \to 0 } \dfrac{1-cos2x}{sen3x}= \lim_{x \to 0 } \dfrac{2sen2x}{3cos3x}=\\\\\\=\dfrac{2.sen(2.0)}{3.cos(3.0)}=\dfrac{2.sen(0)}{3.cos(0)}=\dfrac{2.0}{3.1}=\dfrac{0}{3}=0$

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rubensvlaxius

Limites trigonométricos fundamentais

$\bf{\lim_{x \to 0} \dfrac{ \sin x }{x} = 1}\, \\ e \: \bf{\lim_{x \to 0} \dfrac{ 1 - \cos x }{x} =0}$

$\bf{\lim_{x \to 0} \dfrac{1 - \cos 2x }{ \sin3x}} \\ =\lim_{x \to 0}\bf{\frac{ \frac{2(1 - \cos(2x))}{2x} }{ \frac{ 3\sin(3x) }{3x} }}$

Separando em dois limites temos

$\frac{2\lim_{x \to 0} \frac{1 - \cos(2x) }{2x} }{3\lim_{x \to 0} \frac{ \sin(3x) }{3x} }$

$u=2x\\u\rightarrow\,0\,quando \: x\rightarrow\,0$

$v=3x\\v\rightarrow\,0\,quando \: x\rightarrow\,0$

Substituindo temos :

$\bf\frac{2\lim_{u \to 0} \frac{1 - \cos u }{u} }{3\lim_{v \to 0} \frac{ \sin v}{v} } = \frac{2.0}{3.1} = \bf0$

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Lista de comentários

Limites trigonométricos fundamentais

$\bf{\lim_{x \to 0} \dfrac{ \sin x }{x} = 1}\, \\ e \: \bf{\lim_{x \to 0} \dfrac{ 1 - \cos x }{x} =0}$

$\bf{\lim_{x \to 0} \dfrac{1 - \cos 2x }{ \sin3x}} \\ =\lim_{x \to 0}\bf{\frac{ \frac{2(1 - \cos(2x))}{2x} }{ \frac{ 3\sin(3x) }{3x} }}$

$\frac{2\lim_{x \to 0} \frac{1 - \cos(2x) }{2x} }{3\lim_{x \to 0} \frac{ \sin(3x) }{3x} }$

$\bf\frac{2\lim_{u \to 0} \frac{1 - \cos u }{u} }{3\lim_{v \to 0} \frac{ \sin v}{v} } = \frac{2.0}{3.1} = \bf0$

Me ajudem com esse limite.

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