Question

$\lim_{x \to1} \frac{sen\pi x}{x-1}$

Answer 1

Bom, quando aplicamos na fórmula $x = 1$, chegamos em uma indeterminação.

Aplicamos L'Hôpital:

$f(x) = sen\pi x\\f'(x) = \pi * cos\pi x\\g(x) = x-1\\g'(x) = 1$

$\lim_{x \to 1} \frac{\pi*cos(\pi x) }{1} = \pi*cos(1 * \pi) = \pi*cos(\pi) = -\pi$

Answer 2

$\huge\mathsf{\lim_{x \to1} \frac{sen\pi x}{x-1}}$

Faça

$\mathsf{u=x-1\to~x=u+1}\\\mathsf{x\to~1~quando~x\to~0}$

$\mathsf{\lim_{x \to1} \frac{sen\pi x}{x-1}} = \mathsf{\lim_{u \to0} \frac{sen\pi (u + 1)}{u}}$

$\mathsf{\lim_{u \to0} \frac{sen\pi (u + 1)}{u}} = \mathsf{\lim_{u \to0} \frac{sen (\pi \: u +\pi )}{u}}$

Lembrando que

$\mathsf{\sin(a+b)} \\\mathsf{=\sin(a).\cos(b)+\sin(b).\cos(a)}$

temos

$\mathsf{\sin(\pi~u+\pi)}=\\\mathsf{\sin(\pi~u).\cos(\pi)+\sin(\pi).\cos(\pi~u)}\\\mathsf{\sin(\pi~u).(-1)+0.\cos(\pi~u) } \\\mathsf{ \sin(\pi~u + \pi) = - \sin(\pi~u) }$

Substituindo no limite temos

$\mathsf{\lim_{u \to0} \frac{ \sin(\pi \: u +\pi )}{u}} = \mathsf{\lim_{u \to0} \frac{ - \sin(\pi~u) }{u} }$

Multiplicando e dividindo a expressão por π temos:

$\mathsf{\lim_{u \to0} \frac{ - \sin(\pi~u) }{u} \times \frac{\pi}{\pi}} = \\\mathsf{ - \pi \lim_{u \to0} \frac{ \sin(\pi~u) }{\pi~u}}$

Fazendo

$\mathsf{t=\pi~u}\\\mathsf{t\to~0~quando~u\to~0}$

$\mathsf{ - \pi \lim_{u \to0} \frac{ \sin(\pi~u) }{\pi~u}} = \\ - \mathsf{\pi \lim_{t \to0}\frac{ \sin(t) }{t} } = -\pi.1 = - \pi$