Răspuns:
Explicație pas cu pas:
f(n)=(n+1)e⁻ˣ, f(n+1)=(n+2)e⁻ⁿ⁻¹
[tex]\dfrac{(f(n))^{n}}{e^{n}(f(n+1)^{n}}=(\dfrac{f(n)}{e*f(n+1)} )^{n} =(\dfrac{(n+1)e^{-n}}{e*(n+2)e^{-n-1}})^{n}= (\dfrac{(n+1)e^{-n}}{(n+2)e^{-n}})^{n}=(\dfrac{n+1}{n+2} )^{n}[/tex]
[tex]Deci,~ \lim_{n \to \infty} (\frac{n+1}{n+2})^{n}=( \lim_{n \to \infty} \frac{n(1+\frac{1}{n}) }{n(1+\frac{2}{n} )} )^{n}= ( \lim_{n \to \infty} \frac{1+\frac{1}{n} }{1+\frac{2}{n} } )^{n}=(\frac{1+0}{1+0})^{n}=1^{n}=1.[/tex]
