Răspuns:
Explicație pas cu pas:
[tex]\lim_{n \to \infty} (\frac{n^{2} +n-1}{n^{2}-n+1 } )^{n+3}[/tex]
[tex]\lim_{n \to \infty} (\frac{n^{2} -n+1+2n-2}{n^{2}-n+1 } )^{n+3}=\lim_{n \to \infty} (1+\frac{2n-2}{n^{2} -n+1} )^{n+3}[/tex]
[tex]=\lim_{n \to \infty}[ (1+\frac{2n-2}{n^{2} -n+1} )^{\frac{n^{2}-n+1 }{2n-2} } ]^{\frac{2n-2}{n^{2}-n+1 } } \cdot (n+3)[/tex]
[tex]e^{ \lim_{n \to \infty} \frac{(2n-2)(n+3)}{n^{2} -n+1} }=e^{ \lim_{n \to \infty} \frac{2n^{2}+4n-6 }{n^{2} -n+1} }[/tex]
Avand acelasi grad, adica 2 si la numaratori si la numitor limita va fi 2
deci [tex]e^{2}[/tex] este raspunsul final
