Question

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Answer 1

Răspuns

0

Explicație pas cu pas:

$\lim_{n \to \infty} \frac{e^x}{3^x-1} = (l'Hospital)=\lim_{n \to \infty}\frac{e^x}{3^x ln3} = \frac{1}{ln3}\lim_{n \to \infty} (\frac{e}{3}) ^2$

e=2,71..   => e<3   => $\frac{e}{3}$<1

$\frac{1}{ln3}\lim_{n \to \infty} (\frac{e}{3} )^x =\frac{1}{ln3} * 0 = 0$