Question

calcule os limites limx-0 senx/4x =

Answer 1

Temos o seguinte resultado, chamado de limite fundamental (um deles):

$\boxed{\boxed{\mathsf{\lim\limits_{x\to0}\dfrac{sen\,x}{x}=1}}}$
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Primeira forma: usando o limite fundamental

$\mathsf{\lim\limits_{x\to0}\dfrac{sen\,x}{4x}=\lim\limits_{x\to0}\dfrac{1}{4}\cdot\dfrac{sen\,x}{x}}\\\\\\\mathsf{\lim\limits_{x\to0}\dfrac{sen\,x}{4x}=\dfrac{1}{4}\cdot\lim\limits_{x\to0}\dfrac{sen\,x}{x}}\\\\\\\mathsf{\lim\limits_{x\to0}\dfrac{sen\,x}{4x}=\dfrac{1}{4}\cdot1}\\\\\\\mathsf{\lim\limits_{x\to0}\dfrac{sen\,x}{4x}=\dfrac{1}{4}}$

Segunda forma: usando a regra de L'Hospital

$\mathsf{\lim\limits_{x\to0}\dfrac{sen\,x}{4x}}$

Note que, se $\mathsf{x\to0}$, então $\mathsf{sen\,x\to0}$ e $\mathsf{4x\to0}$, logo temos uma indeterminação do tipo 0/0. Podemos, então, aplicar L'Hospital, já que ambas são funções deriváveis

$\mathsf{\lim\limits_{x\to0}\dfrac{sen\,x}{4x}=\lim\limits_{x\to0}\dfrac{\frac{d}{dx}sen\,x}{\frac{d}{dx}4x}}\\\\\\\mathsf{\lim\limits_{x\to0}\dfrac{sen\,x}{4x}=\lim\limits_{x\to0}\dfrac{cos\,x}{4}}\\\\\\\mathsf{\lim\limits_{x\to0}\dfrac{sen\,x}{4x}=\dfrac{1}{4}\cdot\lim\limits_{x\to0}cos\,x}$

Sabemos que $\mathsf{\lim\limits_{x\to0}cos\,x=1}$, então

$\mathsf{\lim\limits_{x\to0}\dfrac{sen\,x}{4x}=\dfrac{1}{4}\cdot1}\\\\\\\mathsf{\lim\limits_{x\to0}\dfrac{sen\,x}{4x}=\dfrac{1}{4}}$