Question

Calcular $\lim_{x \to \ 1} \frac{sen (x-1)}{(x-1)}$

Answer 1

$\lim\limits_{x\rightarrow1}~\dfrac{sen~(x-1)}{x-1}$

Caso fossemos resolver substituindo x por 1, chegaríamos em sen (0) / 0

Podemos usar a Regra de L'Hopital para resolver limites que geram indeterminações $0/0$ ou $\infty/\infty$

Regra de L'Hopital:

$\boxed{\boxed{\lim\limits_{x\rightarrow a}~\dfrac{f(x)}{h(x)}=\lim\limits_{x\rightarrow a}~\dfrac{f'(x)}{h'(x)}}}$
__________________________

$f(x)=sen~(x-1)\\h(x)=x-1$

Calculando a derivada de f(x):

$f'(x)=\dfrac{d~sen~(x-1)}{dx}=cos~(x-1)*(1x^{1-1}-0)\\\\\\\boxed{f'(x)=cos~(x-1)}$

Calculando a derivada de h(x):

$h'(x)=1x^{1-1}-0\\h'(x)=1$

Logo:

$\lim\limits_{x\rightarrow1}~\dfrac{sen~(x-1)}{x-1}=\lim\limits_{x\rightarrow1}~\dfrac{cos~(x-1)}{1}\\\\\\\lim\limits_{x\rightarrow1}~\dfrac{sen~(x-1)}{x-1}=\lim\limits_{x\rightarrow1}~cos~(x-1)\\\\\\\lim\limits_{x\rightarrow1}~\dfrac{sen~(x-1)}{x-1}=cos~(1-1)\\\\\\\lim\limits_{x\rightarrow1}~\dfrac{sen~(x-1)}{x-1}=cos~0\\\\\\\boxed{\boxed{\lim\limits_{x\rightarrow1}~\dfrac{sen~(x-1)}{x-1}=1}}$