<s>[INST] Limit of $(|x| + |y|)\ln(x^2 + y^4)$ at $(0,0)$ I want to show that $$\lim\limits_{(x,y) \to (0,0)} (\lvert x \rvert + \lvert y \rvert)\ln(x^2 + y^4) = 0$$
First I let $\lVert (x,y) \rVert = \lvert x \rvert + \lvert y \rvert\ < \delta$, and assume that $x,y < 1$ so $x^2 + y^4 < \lvert x \rvert + \lvert y \rvert$. Then $\ln(x^2 + y^4) < \ln(\lvert x \rvert + \lvert y \rvert)$. However, $\lvert \ln(x^2 + y^4)\rvert > \lvert \ln(\lvert x \rvert + \lvert y \rvert)\rvert$, which is where I am stuck because I wanted to show that $\lvert(\lvert x \rvert + \lvert y \rvert)\ln(x^2 + y^4)\rvert < \lvert x \rvert + \lvert y \rvert < \delta $. It does not seem like this approach will work & I am not sure what else I can try.
 [/INST] We have that by generalized mean inequality
$$\frac{\lvert x \rvert + \lvert y \rvert}2 \le \sqrt[4]{\frac{x^4 +y^4}2} \iff \lvert x \rvert + \lvert y \rvert\le\frac 2{\sqrt[4]2}\sqrt[4]{x^4 +y^4}$$
then assuming wlog $x^2 + y^4<1$
$$(\lvert x \rvert + \lvert y \rvert)\left|\ln(x^2 + y^4)\right|
\le \frac 2{\sqrt[4]2}\sqrt[4]{x^4 +y^4}\left|\ln\left(x^4 +y^4\right)\right|= \\
=\frac 8{\sqrt[4]2}\sqrt[4]{x^4 +y^4}\left|\ln\left(\sqrt[4]{x^4 +y^4}\right)\right| 
\to 0$$
since for $t\to 0^+$ we have $t\ln t \to 0$.
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