Question

Aplicando o limite fundamental trigonométrico, resolva:

Answer 1

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$\Large\boxed{\sf{\underline{Limites~trigonom\acute{e}tricos~fundamentais}}}\\\Huge\boxed{\boxed{\boxed{\displaystysle\sf 1)\lim_{x \to 0}\dfrac{sen(x)}{x}=1}}}\\\huge\boxed{\boxed{\boxed{\boxed{\boxed{\displaystyle\sf 2)\lim_{x \to 0}\dfrac{1-cos(x)}{x}=0}}}}}$

$\tt a)~\displaystyle\sf\lim_{x \to 0}\dfrac{sen(5x)}{x}\implies 5\cdot\lim_{x \to 0}\dfrac{sen(5x)}{5x}\\\sf fac_{\!\!,}a~u=5x\\\sf u\longrightarrow0~se~x\longrightarrow 0\\\displaystyle5\cdot\lim_{x \to 0}\dfrac{sen(5x)}{5x}=5\cdot\lim_{u \to 0}\dfrac{sen(u)}{u}=5\cdot1=5\checkmark$

$\tt b)~\displaystyle\sf \lim_{x \to 0}\left(\dfrac{tg(x)}{x}\right)^2\implies\left(\lim_{x \to 0}\dfrac{tg(x)}{x}\right)^2\\\displaystyle\sf\left(\lim_{x \to 0}\dfrac{sen(x)}{x}\cdot\lim_{x \to 0}\dfrac{1}{cos(x)}\right)^2\\\tt separando~os~limites~temos:\\\displaystyle\sf\left(\lim_{x \to}\dfrac{sen(x)}{x}\right)^2\cdot\left(\lim_{x \to 0}\dfrac{1}{cos(x)}\right)^2=1^2\cdot\dfrac{1}{1}=1\checkmark$

$\tt c)\displaystyle\sf\lim_{x \to 0}\dfrac{1-cos(x)}{x}=0\checkmark$