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Given the followning limit: $$ \lim_{x \rightarrow \infty, y \rightarrow \infty} \left( \frac{xy}{x^2 + y^2}\right)^{x^2} $$

To find limit I have made following steps:

  1. Let $ x = y $ ,then limit equals $0$
  2. Let $ x > y $ ,then consider the limit:

$$ \lim_{x \rightarrow \infty, y \rightarrow \infty} \left( \frac{x^2}{x^2 + y^2}\right)^{x^2} = \lim_{x \rightarrow \infty, y \rightarrow \infty} \left( \frac{1}{1 + \frac{y^2}{x^2}}\right)^{x^2} = 0$$ with respect to $$0 < y^2/x^2 < const$$

  1. Let $ y > x $ ,then consider the limit:

$$ \lim_{x \rightarrow \infty, y \rightarrow \infty} \left( \frac{x^2}{x^2 + y^2}\right)^{y^2} = \lim_{x \rightarrow \infty, y \rightarrow \infty} \left( \frac{1}{\frac{x^2}{y^2} + 1}\right)^{x^2} = 0$$ with respect to $$0 < x^2/y^2 < const$$

What could you say about my solution?

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  • $\begingroup$ Instead of $x=y$ you should say $\displaystyle\lim\left(\frac xy\right) = 1$. $\endgroup$ – user202729 Sep 24 '16 at 15:30
  • $\begingroup$ And for another cases: $0 < lim(x/y) < 1$ and $1 < lim(x/y) < const$? $\endgroup$ – eaniconer Sep 24 '16 at 15:42
  • $\begingroup$ "What could you say about my solution?" That it does not suffice to solve the question. $\endgroup$ – Did Sep 24 '16 at 16:17
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For $x, y >1$, we have the fact that \begin{align} 2\leq\frac{x}{y}+\frac{y}{x} \end{align} which means \begin{align} \left(\frac{xy}{x^2+y^2}\right)^{x^2}=\left(\frac{1}{\frac{x}{y}+\frac{y}{x}}\right)^{x^2} \leq \left(\frac{1}{2}\right)^{x^2}. \end{align}

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  • $\begingroup$ Good fact, thank.. $\endgroup$ – eaniconer Sep 24 '16 at 18:54

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