Răspuns:
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Explicație pas cu pas:
Aplicam formula [tex]a^{log_a~b}=b[/tex], dar si proprietatea de la logaritmi care ne spune ca [tex]log_a b^c=c*log_ab[/tex].
Totodata, stim si ca [tex]lim_{n \to \infty} e^{a_n}=e^{lim_{n \to \infty} a_n}[/tex], aₙ fiind un sir oarecare.
[tex]L=\lim_{n \to \infty} (\frac{n}{7n+3})^n= \lim_{n \to \infty} e^{ln(\frac{n}{7n+3})^n}= \lim_{n \to \infty} e^{nln(\frac{n}{7n+3})}=e^{ \lim_{n \to \infty} nln(\frac{n}{7n+3})}[/tex]
Calculam separat limita ramasa la exponent.
Aplicam, totodata, si formula: [tex]lim_{n \to \infty} a_n*b_n=lim_{n \to \infty} a_n*lim_{n \to \infty}b_n[/tex], cand sirurile sunt finite (au numar finit de termeni).
[tex]{ \lim_{n \to \infty} nln(\frac{n}{7n+3})}=lim_{n \to \infty}n*lim_{n \to \infty}ln(\frac{n}{7n+3})=\infty*lim_{n \to \infty} ln(\frac{n}{n(7+\frac{3}{n})})=\infty*ln(\frac{1}{7})=\infty*(-ln7)=-\infty[/tex]
Si ne intoarcem la limita noastra:
[tex]L=lim_{n \to \infty} (\frac{n}{7n+3})^n=e^{ \lim_{n \to \infty} nln(\frac{n}{7n+3})}=e^{-\infty}=\frac{1}{e^{\infty}}=0[/tex]
