Question

Admitere automatica Iasi 2018 Gheorghe Asachi

Answer 1

Răspuns:

d)  $\sqrt{2}$

Explicație pas cu pas:

$(x_{n}),(y_{n} )$ doua siruri de numere rationale.

Consideram $a_{n} =(1+\sqrt{2}) ^{n} =x_{n} +y_{n} \sqrt{2}$

si $b_{n} =(1-\sqrt{2})^n=x_{n} -y_{n} \sqrt{2}$

$\left \{ {{a_{n} =x_{n}+y_{n}\sqrt{2}  } \atop {b_{n} =x_{n} -y_{n}\sqrt{2}  }} \right.$

$a_{n} +b_{n} =2x_{n} =>x_{n} =\frac{a_{n}+b_{n}  }{2}$

$a_{n} -b_{n} =2\sqrt{2} y_{n} =>y_{n} =\frac{a_{n}-b_{n}  }{2\sqrt{2} }$

$\lim_{n \to \infty} \frac{x_{n} }{y_{n} } = \lim_{n \to \infty} \frac{\frac{a_{n} +b_{n} }{2} }{\frac{a_{n}-b_{n}  }{2\sqrt{2} } } } = \lim_{n \to \infty} \frac{\sqrt{2} (a_{n}+b_{n})  }{a_{n}-b_{n} }$

$\lim_{n \to \infty} \frac{\sqrt{2}[(\sqrt{2}+1)^n+(1-\sqrt{2})^n]  }{[(\sqrt{2}+1)^n-(1-\sqrt{2}^n]  } =\frac{\sqrt{2}(\sqrt{2}+1)^n[1+\frac{(1-\sqrt{2)}^n }{(\sqrt{2}+1)^n} ]}{(\sqrt{2} +1)^n[1-\frac{(1-\sqrt{2})^n }{(1+\sqrt{2})^n}]}=\frac{\sqrt{2}(1+0) }{(1-0)} =\sqrt{2}$