I am asked to prove that the limit of this multivariable function is equal to zero. I used the squeeze theorem to say that the limit is greater than 0 and less than x+y, by taking the limit of those we get that it is between zero and zero thus making it = to 0.
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Did you mean to write $\sqrt{|x-y|}$ by any chance, or perhaps the "one-sided" limit where both $x$ and $y$ approach $0$ from the "right"? – user170231 Mar 03 '16 at 18:42
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As stated, we need $x\ge y,x+y\ne 0.$ – zhw. Mar 03 '16 at 19:08
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The limit is not $0$ and in fact does not exist.
If the limit is taken along the line $x=y$, then the limit is trivially $0$.
However, if the limit is approached along the path described parametrically by $x(t)=t$ and $y(t)=-t+e^{-1/t}$, then the limit becomes
$$\begin{align}\lim_{(x,y)\to (0,0)}\sqrt{|x-y|}\log|x+y|&=-\lim_{t\to 0}\frac{\sqrt{|2t-e^{-1/t}|}}{t}\\\\&=-\infty\end{align}$$
Mark Viola
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There may be some lack of rigor in this, but still.
$$\sqrt{x-y}\log(|x+y|) = \log\left(|x+y|^{\sqrt{x-y}}\right)$$
Since $x\to 0$ and $y\to 0$ we can say either $(x+y)\to 0$ and $(x-y)\to 0$.
Now we know that
$$\lim_{z\to 0}\ z^z = 1$$
And also
$$\lim_{z\to 0}\ z^{\sqrt{z}} = 1$$ and so
$$\log\left(|x+y|^{\sqrt{x-y}}\right) = \log(1) = 0$$
The lack of rigor consists in the fact that actually we may have $x-y = i\epsilon$ for $|\epsilon| <<< 1$.
Enrico M.
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$|x+y| \neq \sqrt{x-y}$, show how can you use $\lim_{z \to 0} z^z=1$? – Hetebrij Mar 03 '16 at 19:04
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$$\lim_{x\to 0} x^{\sqrt{x}} = 1$$ holds again.
I assumed that $x+y$ and $x-y$ were about similar, but again: the problem that could arise is written in the end of the answer
– Enrico M. Mar 03 '16 at 19:14 -
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That is true @Hetebrij Indeed I played a bit, it may have been useful somehow. – Enrico M. Mar 03 '16 at 19:16