Question

limite de (-2x+3sin+pi-3)/(x-pi/2) quando o x tender para pi/2

Answer 1

$\lim_{x \to \frac{\pi}{2} } \frac{-2x \ + \ 3sin(x) \ + \ \pi \ - \ 3}{x \ - \ \frac{\pi}{2}} =$
$\lim_{x \to \frac{\pi}{2} } \frac{(-2x \ + \ \pi)}{x \ - \frac{\pi}{2}} \ + \ \frac{ 3sin(x) \ - \ 3}{x \ - \ \frac{\pi}{2}} =$
$\lim_{x \to \frac{\pi}{2} } \frac{-2(x \ - \ \frac{\pi}{2})}{x \ - \frac{\pi}{2}} \ + \ \frac{ 3(sin(x) \ - \ 1)}{x \ - \ \frac{\pi}{2}}$

$Fazendo \ u = x - \frac{\pi}{2}$
$\lim_{x \to \frac{\pi}{2} } -2 \ + \ \frac{ 3(sin(x) \ - \ 1)}{x \ - \ \frac{\pi}{2}} =$
$\lim_{u \to 0 } -2 \ + \ \frac{ 3(sin(u+\frac{\pi}{2}) \ - \ 1)}{u}$

$Como \ sin(a+b) =sin(a)cos(b) + sin(b)cos(a)$
$sin (u+\frac{\pi}{2})=cos(u)$
$\lim_{u \to 0 } -2 \ + \ \frac{ 3(sin(u+\frac{\pi}{2}) \ - \ 1)}{u} =$
$\lim_{u \to 0 } -2 \ + \ \frac{ 3(cos(u) \ - \ 1)}{u} =$
$\lim_{u \to 0 } -2 \ + \ \frac{ 3(cos(u) \ - \ 1)}{u} \frac{(cos(u) \ + \ 1)}{(cos(u) \ + \ 1)}=$
$-2 \ + \ 3\lim_{u \to 0 } \frac{sin^{2}(u)}{u(cos(u) \ - \ 1)} =$
$-2 \ + \ 3\lim_{u \to 0 } \frac{sin(u)}{u} \frac{sin(u)}{(cos(u) \ - \ 1)}$

Como $\lim_{u \to 0} \frac{sin(u)}{u} = 1$
E $\lim_{u \to 0} \frac{sin(u)}{cos(u) \ + \ 1} = \frac {0}{1 \ + \ 1} = 0$

então, $-2 \ + \ 3\lim_{u \to 0 } \frac{sin(u)}{u} \frac{sin(u)}{(cos(u) \ - \ 1)} = -2$


Logo, $\boxed{\boxed{\lim_{x \to \frac{\pi}{2} } \frac{-2x \ + \ 3sin(x) \ + \ \pi \ - \ 3}{x \ - \ \frac{\pi}{2}} =-2}}$