I want to calculate limit of $\lim _{\left(x,y\right)\to \left(0,0\right)}\left(\left(xy\right)\ln \left(x^2+y^2\right)\right)$ using Squeeze theorem or using definition of limit. please help
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Some context or your own attempt would help us help you. Thanks – Benjamin Wang Jul 18 '20 at 15:37
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2Does this answer your question? Showing $\lim_{(x,y) \to (0,0)} xy \log(x^2+y^2) = 0$ – Souza Jul 18 '20 at 15:38
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HINT:
Note that $|xy|\le \frac12(x^2+y^2)$ so that
$$\left|xy\log(x^2+y^2)\right|\le (x^2+y^2)\left|\log\left(\sqrt{x^2+y^2}\right)\right|$$
Can you finish now?
Mark Viola
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$|\ln{\sqrt{x^2+y^2}}|$ is defined when $(x,y) \to (0,0)$ ? How do I do then? – Jhon 117 Jul 18 '20 at 16:22
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Note that $\log(x)\le \sqrt{x}$. So for $x<1$, $|\log(x)|\le \frac{1}{\sqrt{x}}$. Now just replace $x$ with $x^2+y^2$. – Mark Viola Jul 18 '20 at 17:45
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thanks, i still don't understand, $\frac{1}{\sqrt{x}}$ not defined when $x \to 0$ – Jhon 117 Jul 18 '20 at 17:53
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You're right. $\lim_{x\to0^+}\frac1{\sqrt{x}}$ is not defined. But $\lim_{x\to0} x/\sqrt{x}=0$ is defined. ;-) – Mark Viola Jul 18 '20 at 18:16
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$|(x^2+y^2)\log(\sqrt{x^2+y^2})|\le \frac{x^2+y^2}{2\sqrt{x^2+y^2}}=\frac12\sqrt{x^2+y^2}$ – Mark Viola Jul 18 '20 at 19:09