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.

I will use the following inequality in this answer :
$\ln z<z\;\;$ for any $\;z\in\Bbb R^+.\quad\color{blue}{(*)}$
For any $\,(x,y)\in\Bbb R^2\!\setminus\!\{(0,0)\}\,$ such that $\,x^2+y^2<1\,,\,$ it results that
$\begin{align}\color{blue}{-16\sqrt[4]{x^2+y^2}}&=-8\sqrt[4]{x^2+y^2}-8\sqrt[4]{x^2+y^2}\leqslant\\&\leqslant-8\sqrt[4]{x^2}-8\sqrt[4]{y^2}=-8\sqrt[8]{x^4}-8\sqrt[8]{y^4}\leqslant\\&\leqslant-8\sqrt[8]{\dfrac{x^8}{x^4+y^4}}-8\sqrt[8]{\dfrac{y^8}{x^4+y^4}}=\\&=-8\dfrac{|x|}{\sqrt[8]{x^4+y^4}}-8\dfrac{|y|}{\sqrt[8]{x^4+y^4}}=\\&=-8\big(|x|+|y|\big)\!\cdot\!\dfrac1{\sqrt[8]{x^4+y^4}}<\\&\overset{\color{brown}{(*)}}{<}-8\big(|x|+|y|\big)\ln\!\bigg(\!\dfrac1{\sqrt[8]{x^4+y^4}}\!\!\bigg)=\\&=\big(|x|+|y|\big)\ln\left(x^4+y^4\right)<\\&\color{blue}{<\big(|x|+|y|\big)\ln\left(x^2+y^4\right)}<\\&<\big(|x|+|y|\big)\ln\left(x^2+y^2\right)\color{blue}{<0}\;.\end{align}$
Since $\!\lim\limits_{(x,y)\to(0,0)}\!\left(-16\sqrt[4]{x^2+y^2}\right)=0\;,\;$ by applying the Squeeze theorem, it follows that
$\lim\limits_{(x,y)\to(0,0)}\!\big(|x|+|y|\big)\ln\left(x^2+y^4\right)=0\;.$
