# Limit of two variables $\lim_{x \rightarrow \infty, y \rightarrow \infty} \left( \frac{xy}{x^2 + y^2}\right)^{x^2}$

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?

• Instead of $x=y$ you should say $\displaystyle\lim\left(\frac xy\right) = 1$. – user202729 Sep 24 '16 at 15:30
• And for another cases: $0 < lim(x/y) < 1$ and $1 < lim(x/y) < const$? – eaniconer Sep 24 '16 at 15:42
• "What could you say about my solution?" That it does not suffice to solve the question. – Did Sep 24 '16 at 16:17

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}