Question

Calcule (sem utilizar a regra de l'hôpital)


 lim_{x \to \infty + } (0,2+0.3 e^{0.02x} )[/tex]

Answer 1

$\lim_{x \to +\infty}\left(0,2+0,3e^{0,02x} \right )\\ \\ =\lim_{x \to +\infty}\left(0,2\right)+\lim_{x \to +\infty}\left( 0,3e^{0,02x} \right )\\ \\ =0,2+0,3\underbrace{\lim_{x \to +\infty}e^{0,02x}}_{+\infty}\\ \\ =+\infty$


$\boxed{\lim_{x \to +\infty}\left(0,2+0,3e^{0,02x} \right )=+\infty}$