Răspuns:
1
Explicație pas cu pas:
[tex]\lim_{n \to \infty} (\frac{n^{2}-3n+2 }{n^{2}+n+1 })^{\sqrt{n} } = \lim_{n \to \infty} (1+\frac{n^{2}-3n+2 }{n^{2}+n+1 }-1)^{\sqrt{n} } = \lim_{n \to \infty} (1+\frac{n^{2}-3n+2-n^{2}-n-1 }{n^{2}+n+1 })^{\sqrt{n} } = \lim_{n \to \infty}(1+\frac{-4n+1}{n^{2}+n+1 })^{\sqrt{n} } = \lim_{n \to \infty} ((1+\frac{-4n+1}{n^{2}+n+1 })^{\frac{n^{2}+n+1 }{-4n+1} })^{\frac{-4n+1}{n^{2}+n+1 } \sqrt{n} })=e^{0}=1[/tex]
