Răspuns:
d) [tex]\sqrt{2}[/tex]
Explicație pas cu pas:
[tex](x_{n}),(y_{n} )[/tex] doua siruri de numere rationale.
Consideram [tex]a_{n} =(1+\sqrt{2}) ^{n} =x_{n} +y_{n} \sqrt{2}[/tex]
si [tex]b_{n} =(1-\sqrt{2})^n=x_{n} -y_{n} \sqrt{2}[/tex]
[tex]\left \{ {{a_{n} =x_{n}+y_{n}\sqrt{2} } \atop {b_{n} =x_{n} -y_{n}\sqrt{2} }} \right.[/tex]
[tex]a_{n} +b_{n} =2x_{n} =>x_{n} =\frac{a_{n}+b_{n} }{2}[/tex]
[tex]a_{n} -b_{n} =2\sqrt{2} y_{n} =>y_{n} =\frac{a_{n}-b_{n} }{2\sqrt{2} }[/tex]
[tex]\lim_{n \to \infty} \frac{x_{n} }{y_{n} } = \lim_{n \to \infty} \frac{\frac{a_{n} +b_{n} }{2} }{\frac{a_{n}-b_{n} }{2\sqrt{2} } } } = \lim_{n \to \infty} \frac{\sqrt{2} (a_{n}+b_{n}) }{a_{n}-b_{n} }[/tex]
[tex]\lim_{n \to \infty} \frac{\sqrt{2}[(\sqrt{2}+1)^n+(1-\sqrt{2})^n] }{[(\sqrt{2}+1)^n-(1-\sqrt{2}^n] } =\frac{\sqrt{2}(\sqrt{2}+1)^n[1+\frac{(1-\sqrt{2)}^n }{(\sqrt{2}+1)^n} ]}{(\sqrt{2} +1)^n[1-\frac{(1-\sqrt{2})^n }{(1+\sqrt{2})^n}]}=\frac{\sqrt{2}(1+0) }{(1-0)} =\sqrt{2}[/tex]
