Question

Qual o valor do limite? $\lim_{x \to \frac{3 \pi }{2} } [cos( \frac{ \pi }{2}-x)]$
Opções:
a) 2
b) 1
c) 3
d) -5
e) 0

Answer 1

Teorema

Sejam f(x) e g(x) duas funções

Se $\lim\limits_{x\rightarrow a}f(x)=L$ e g(x) é contínua em L, então

$\boxed{\boxed{\lim\limits_{x\rightarrow a}g(f(x))=g\left(\lim\limits_{x\rightarrow a}f(x)\right)=g(L)}}$
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A função cos(x) é contínua para todo x pertencente aos reais, logo

$\lim\limits_{x\rightarrow(\frac{3\pi}{2})}cos\left(\dfrac{\pi}{2}-x\right)=cos\left(\lim\limits_{x\rightarrow(\frac{3\pi}{2})}\left[\dfrac{\pi}{2}-x}\right]\right)\\\\\\\lim\limits_{x\rightarrow(\frac{3\pi}{2})}cos\left(\dfrac{\pi}{2}-x\right)=cos\left(\lim\limits_{x\rightarrow(\frac{3\pi}{2})}\dfrac{\pi}{2}-\lim\limits_{x\rightarrow(\frac{3\pi}{2})}x\right)\\\\\\\lim\limits_{x\rightarrow(\frac{3\pi}{2})}cos\left(\dfrac{\pi}{2}-x\right)=cos\left(\dfrac{\pi}{2}-\dfrac{3\pi}{2}\right)$

$\lim\limits_{x\rightarrow(\frac{3\pi}{2})}cos\left(\dfrac{\pi}{2}-x\right)=cos\left(\dfrac{-2\pi}{2}\right)\\\\\\\lim\limits_{x\rightarrow(\frac{3\pi}{2})}cos\left(\dfrac{\pi}{2}-x\right)=cos(-\pi)$

Sabemos que

$cos(-x)=cos(0-x)=cos(0)cos(x)+sen(0)sen(x)=cos(x)$

Logo:

$\lim\limits_{x\rightarrow(\frac{3\pi}{2})}cos\left(\dfrac{\pi}{2}-x\right)=cos(-\pi)=cos(\pi)\\\\\\\boxed{\boxed{\lim\limits_{x\rightarrow(\frac{3\pi}{2})}cos\left(\dfrac{\pi}{2}-x\right)=-1}}$