Question

Calcule (sem utilizar a regra de l'hôpital)

$\lim_{x \to \ 0+} \frac{log (base3) x}{ e^{x}-1 }$

Answer 1

$\lim_{x \to 0^{+}}\frac{\mathrm{\ell og}_{3}x}{e^{x}-1}\\ \\ =\lim_{x \to 0^{+}}\left(\mathrm{\ell og}_{3}\,x \cdot \frac{1}{e^{x}-1} \right )\\ \\ =\underbrace{\lim_{x \to 0^{+}}\mathrm{\ell og}_{3}\,x}_{-\infty} \cdot \underbrace{\lim_{x \to 0^{+}}\frac{1}{e^{x}-1}}_{+\infty}\\ \\ =-\infty$


$\boxed{\lim_{x \to 0^{+}}\frac{\mathrm{\ell og}_{3}x}{e^{x}-1}=-\infty}$