Răspuns:
[tex]e^\frac{5}{7}[/tex]
Explicație pas cu pas:
[tex]\lim_{n \to \infty} (\frac{7n+4}{7n-1} )^{\frac{n^2+n}{n} }= \lim_{n \to \infty} (\frac{7n-1 +5}{7n-1} )^{\frac{n^2+n}{n} } \\\\ \lim_{n \to \infty} (\frac{7n-1}{7n-1}+\frac{5}{7n-1} )^{\frac{n^2+n}{n} }= \lim_{n \to \infty} (1+\frac{5}{7n-1} )^{\frac{n^2+n}{n} }\\\\ \lim_{n \to \infty} [(1+\frac{5}{7n-1} )^{\frac{7n-1}{5} }]^{\frac{n^2+n}{n} *\frac{5}{7n-1}} = e^ \lim_{n \to \infty} \frac{n^2+n}{n} *\frac{5}{7n-1} } \\\\e^\lim_{n \to \infty} \frac{5n^2+5n}{7n^2-n}}=e^{\frac{5}{7} }[/tex]
