Question

Calcule o seguinte limite:
$\lim_{x \to \0} \frac{1-cos(3x)}{x^{2} }$
Obs: quando x se aproxima de 0 (não consegui colocar na fórmula)

Answer 1

Oie, tudo bom?

$= lim_{x⟶0}( \frac{1 - \cos(3x) }{x {}^{2} } )$

Dado que avaliar os limites do numerador e do denominador resultaria numa fórmula indeterminada, use a regra de L'Hopital:

$lim_{x⟶c}( \frac{f(x)}{g(x)} ) = lim_{x⟶c}( \frac{f'(x)}{g'(x)} )$

$= lim_{x⟶0}( \frac{ \frac{d}{dx}(1 - \cos(3x) ) }{ \frac{d}{dx} (x {}^{2} )} )$

$= lim_{x⟶0}( \frac{ \frac{d}{dx} (1) - \frac{d}{dx} ( \cos(3x)) }{2x {}^{2 - 1} } )$

$= lim_{x⟶0}( \frac{0 - ( - \sin(3x) \: . \: 3)}{2x {}^{1} } )$

$= lim_{x⟶0}( \frac{3 \sin(3x) }{2x} )$

Dado que avaliar os limites do numerador e do denominador resultaria numa fórmula indeterminada, use a regra de L'Hopital:

$l im_{x⟶c}( \frac{f(x)}{g(x)} ) = log_{x⟶c}( \frac{f'(x)}{g'(x)} )$

$= lim_{x⟶0}( \frac{ \frac{d}{dx} (3 \sin(3x)) }{ \frac{d}{dx}(2x) } )$

$= lim_{x⟶0}( \frac{3 \: . \: \frac{d}{dx} ( \sin(3x)) }{2} )$

$= lim_{x⟶0}( \frac{3 \: . \: \frac{d}{dg}( \sin(g) ) \: . \: \frac{d}{dx} (3x)}{2} )$

$= lim_{x⟶0}( \frac{3 \cos(g) \: . \: 3 }{2} )$

$= lim_{x⟶0}( \frac{3 \cos(3x) \: . \: 3 }{2} )$

$= lim_{x⟶0}( \frac{9 \cos(3x) }{2} )$

$= \frac{ lim_{x⟶0}(9 \cos(3x) ) }{ log_{x⟶0}(2) }$

$= \frac{9 \: . \: lim_{x⟶0}( \cos(3x) ) }{2}$

$= \frac{9 \cos( lim_{x⟶0}(3x) ) }{2}$

$= \frac{9 \cos(3 \: . \: lim_{x⟶0}(x) ) }{2}$

$= \frac{9 \cos(3 \: . \: 0) }{2}$

$= \frac{9 \cos(0) }{2}$

$= \frac{9 \: . \: 1}{2}$

$\boxed{= \frac{9}{2} }$

Att. NLE Top Shotta