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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?

jimjim
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eaniconer
  • 67

2 Answers2

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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}

Jacky Chong
  • 25,739
  • Good fact, thank.. – eaniconer Sep 24 '16 at 18:54
  • Do you need $x,y > 1$ or just $x,y > 0$? It seems to follow from the square of $x-y$ is non-negative so $0 \leq (x-y)^2 = x^2 + y^2 - 2xy$. In order to solve for $2$ we'll need to multiply by $1/y$ and $/1/x$ which should be fine if they're greater than zero. – CyclotomicField Apr 29 '20 at 20:52
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Hint,

Assume $x = r \cos \theta$ and $y = r \sin \theta, \theta = constant$

Then, the limit changes to

$$ \lim \limits_{r \rightarrow \infty}\left ( \dfrac {r^2 \sin \theta \cos \theta}{r^2}\right)^{r^2 \cos^2 \theta} \\ \Rightarrow \lim \limits_{r \rightarrow \infty} (\sin \theta\cos \theta)^{r^2 \cos^2\theta} = 0$$

As expression under bracket is $<1$