Question

Socorrooooooooooooooo

Answer 1

Aplicando a Regra de L'Hôpital:

$\lim_{x \rightarrow -4} \dfrac{3^{\frac{x+4}{7}}}{sen\left[5 \cdot (x+4) \right]} = \lim_{x \rightarrow -4} \dfrac{1}{7} \cdot \dfrac{3^{\frac{x+4}{7}}\cdot \ln[3]}{cos\left[5 \cdot (x+4) \right] \cdot 5}$

Aplicando o Limite:

$\lim_{x \rightarrow -4} \dfrac{3^{\frac{x+4}{7}}}{sen\left[5 \cdot (x+4) \right]} = \dfrac{1}{7} \cdot \dfrac{3^{\frac{-4+4}{7}}\cdot \ln[3]}{cos\left[5 \cdot (-4+4) \right] \cdot 5}$

$\lim_{x \rightarrow -4} \dfrac{3^{\frac{x+4}{7}}}{sen\left[5 \cdot (x+4) \right]} = \dfrac{1}{7} \cdot \dfrac{3^{0}\cdot \ln[3]}{cos\left[0\right] \cdot 5}$

$\lim_{x \rightarrow -4} \dfrac{3^{\frac{x+4}{7}}}{sen\left[5 \cdot (x+4) \right]} = \dfrac{1}{7} \cdot \dfrac{1 \cdot \ln[3]}{1 \cdot 5}$

$\lim_{x \rightarrow -4} \dfrac{3^{\frac{x+4}{7}}}{sen\left[5 \cdot (x+4) \right]} = \dfrac{1}{7} \cdot \dfrac{\ln[3]}{5}$

$\lim_{x \rightarrow -4} \dfrac{3^{\frac{x+4}{7}}}{sen\left[5 \cdot (x+4) \right]} = \dfrac{\ln[3]}{35}$

Alternativa A