Răspuns
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Explicație pas cu pas:
[tex]\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\\[/tex]
e=2,71.. => e<3 => [tex]\frac{e}{3}[/tex]<1
[tex]\frac{1}{ln3}\lim_{n \to \infty} (\frac{e}{3} )^x =\frac{1}{ln3} * 0 = 0[/tex]
