Let $x_{n}$, $y_{n}\ge 0$ satisfy the equations

$$\large\begin{cases} x^{\frac{4n+5}{n}}_{n}+3x_{n}+y_{n}=1+\dfrac{1}{n}\\ y^{\frac{n+2}{n}}_{n}+4x_{n}+3y_{n}=4-\dfrac{1}{n} \end{cases}$$

Show that the limits


exist, and find their values.


I call your two relations (1) and (2). Let us show first that $(x_n) \to 0$. It will suffice to show that for any $n$, $$ x_n \leq \frac{4}{5n} \tag{*} $$

If $x_n \le \frac{1}{3n}$ then (*) is obvious. So assume $x_n \ge \frac{1}{3n}$. We then have, by (1) $$ y_n=1+\frac{1}{n}-x_n^{\frac{4n+5}{n}}-3x_n \leq 1+\frac{1}{n}-3x_n \leq 1 \tag{3} $$

Now (2) forces us to have

$$ (4-\frac{1}{n}-4x_n-3y_n)^n=y_n^{n+2} \leq 1 \tag{4} $$

We deduce

$$ 1 \geq 4-\frac{1}{n}-4x_n-3y_n=4-\frac{1}{n}-4x_n-3\big(1+\frac{1}{n}-x_n^{\frac{4n+5}{n}}-3x_n)\big)=1-\frac{4}{n}-5x_n-3x_n^{\frac{4n+5}{n}} \tag{5} $$

and hence $$ 5x_n+3x_n^{\frac{4n+5}{n}} \leq \frac{4}{n} \tag{6} $$ This proves (*).

Now we know that $x_n \to 0$, we deduce $y_n \to 1$ by (1), and this finishes the problem.

  • $\begingroup$ If $(x_n, y_n)$ exists! $\endgroup$ – zyx Sep 28 '13 at 23:50
  • $\begingroup$ One can show $(x_n,y_n)$ exists, but that’s a separate question in my view. $\endgroup$ – Ewan Delanoy Sep 29 '13 at 2:31

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