Răspuns:
Explicație pas cu pas:
[tex]\lim_{n \to \infty} \frac{3\cdot 2^n+2\cdot 3^n^+^1}{5\cdot 4^n^-^1-4\cdot 5^n} \\\\[/tex]
dam factor comun
[tex]\lim_{n \to \infty} \frac{3^n^+^1(\frac{3\cdot2^n}{3\cdot 3^n}+2) }{5^n(\frac{5\cdot 4^n^-^1}{5\cdot5^n^-^1}-4) }[/tex]
se simplifica 3 cu 3 si 5 cu 5
[tex]\lim_{n \to \infty} \frac{3^n^+^1(\frac{2^n}{ 3^n}+2) }{5^n(\frac{4^n^-^1}{5^n^-^1}-4) }[/tex]
[tex]\lim_{n \to \infty} \frac{3^n^+^1[(\frac{2}{ 3})^n+2] }{5^n[(\frac{4}{5})^n^-^1-4]}[/tex]
orice fractie subunitara la puterea ∞ tinde catre 0
adica [tex](\frac{2}{3} )^n, (\frac{4}{5} )^n^-^1[/tex] tind catre 0
Deci ne ramane
[tex]\lim_{n \to \infty} \frac{3^n\cdot 3\cdot 2}{5^n\cdot(-4)} = \lim_{n \to \infty} (\frac{3}{5} )^n\cdot (\frac{3}{-2} )[/tex]=0
pentru ca [tex](\frac{3}{5} )^n[/tex] tinde catre 0, fiind subunitara
Sper ca ai inteles!
