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I have a question on the limit of $$\lim_{x,y\to\infty}\frac{(x-1)(y-1)}{xy}$$ I had a look on answers and theory like the following question: Limit question as $x$ and $y$ approach infinity?

So if I'm getting it right, the limit must exist by approaching by any path, that is, if we make $y=x$

$$\lim_{x\to\infty}\frac{(x-1)^2}{x^2}=1$$

which also holds for $y=x^2$, but not for things like $y=x^{-2}$:

$$\lim_{x,y\to\infty}{x(x-1)(x^{-2}-1)}=-\infty$$

and thus the limit doesn't exist. Am I getting it right?

Thanks for your help!

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    $y=x^{-2}$ does not describe a path along which *both* $x,y\to \infty$ *simultaneously*, so no – 2017-01-27
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    Ohh! I get it, the path must lead both simultaneusly to infinity! That wasn't clear for me until now... thanks! – 2017-01-27
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    glad to be of help! – 2017-01-27

3 Answers 3

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It is unclear exactly what $\lim\limits_{x,y\to\infty}f(x,y)$ means.

Your comments are correct for $$ \lim_{x^2+y^2\to\infty}f(x,y) $$ However, one might also consider the limit $$ \lim_{\min(x,y)\to\infty}f(x,y) $$ for which your argument does not work.

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HINT:

$$\lim_{x,y\to\infty}\frac{(x-1)(y-1)}{xy} = \lim_{x,y\to\infty}{(1-\frac1x)(1-\frac1y)}{}$$

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    i think author is interested not in the answer, but he wants to know if his method is correct – 2017-01-27
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    @MarkoffChainz I commented on that point. You might want to expand on that – 2017-01-27
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    exactly, so then would it be ritght to say $$\lim_{x,y\to\infty}\frac{(x-1)(y-1)}{xy} = \lim_{x,y\to\infty}{(1-\frac1x)(1-\frac1y)}{}= \lim_{x\to\infty}{(1-\frac1x)\lim_{y\to\infty}{(1-\frac1y)}}{}=1$$? – 2017-01-27
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se that $$\frac{(x-1)(y-1)}{xy}=\frac{xy-y-x+1}{xy}=1-\frac{1}{x}-\frac{1}{y}+\frac{1}{xy}$$