<s>[INST] Limit of ${ \lim_{(x,y)\to(0,0)} {(\left| x \right| + \left| y \right|) \ln{(x^2 + y^4)} }}$ This question comes from Hubbard's textbook "Vector Calculus, Linear Algebra, and Differential Forms: A Unified Approach" (5th edition, exercise 1.5.16b)
I wrote a $\epsilon -\delta$ proof to show ${ \lim_{(x,y)\to(0,0)} {(\left| x \right| + \left| y \right|) \ln{(x^2 + y^4)} }} = 0$, but wanted to ask to check if what I wrote was correct:

Suppose we have $\epsilon > 0$.

*

*First, it can be shown (with L'Hoptial for example) that $\lim_{x\to 0} |x| \ln(x^k) = 0$ for even positive integers $k$. Thus there exists $\delta_{1;k} > 0$ such that $|x| < \delta_{1;k}$ implies $\left| |x| \ln(x^k) \right| = \left| k |x| \ln(x) \right| < \frac{k\epsilon}{6}$. Therefore: $|x| \ln(x) > - \frac{\epsilon}{6}$.


*Secondly, clearly we have $\lim_{x \to 0} {|x| \ln(x^2 + c)} = \lim_{y \to 0} {|y| \ln(y^4 + c)} = 0$ for any $c > 0$. So there exists $\delta_{2; c}, \delta_{3;c} > 0$ such that:

*

*$|x| < \delta_{2;c} \implies \left| |x| \ln(x^2 + c) \right| < \frac{\epsilon}{2}$.

*$|y| < \delta_{3;c} \implies \left| |y| \ln(y^4 + c) \right| < \frac{\epsilon}{2}$.




Let $\delta = \min(\epsilon, \delta_{1;2}, \delta_{1;4}, \delta_{2;\epsilon}, \delta_{3;\epsilon})$. So if $\sqrt{x^2 + y^2} < \delta$, note that this implies $|x| < \delta$ and $|y| < \delta$. Then:
$$\begin{align} (|x| + |y|) \ln(x^2 + y^4) &= |x| \ln(x^2 + y^4) + |y| \ln(x^2 + y^4) \\ &\leq |x| \ln(x^2 + \epsilon^4) + |y| \ln(\epsilon^2 + y^4) \\ &< \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \end{align}$$
As $x^2 + y^4 < \epsilon^2 + y^4$ (as $|x| < \delta \leq \epsilon$) and $\ln(x)$ is a strictly increasing implies $\ln(x^2 + y^4) < \ln(\epsilon^2 + y^4)$. Since $|y| \geq 0$ then $|y| \ln(x^2 + y^4) \leq |y| \ln(\epsilon^2 + y^4)$.
Moreover:
$$\begin{align} (|x| + |y|) \ln(x^2 + y^4) &= |x| \ln(x^2 + y^4) + |y| \ln(x^2 + y^4)
\\ &\geq |x| \ln(x^2) + |y| \ln(y^4) \\ &= 2|x|\ln(x) + 4|y| \ln(y) \\ &> -\frac{2\epsilon}{6} - \frac{4\epsilon}{6} = -\epsilon \end{align}$$
Putting these together we have $\left| (|x| + |y|) \ln(x^2 + y^4) \right| < \epsilon$.
Therefore  ${ \lim_{(x,y)\to(0,0)} {(\left| x \right| + \left| y \right|) \ln{(x^2 + y^4)} }} = 0$.
 [/INST] Given that the function is even with respect to $x$ and with respect to $y,$ it is enough to consider $x,y\geq0,$ and $(x,y)\neq(0,0).$  I would write
$$
\lim_{(x,y)\to(0,0)}\frac{x+y}{\sqrt[4]{x^2+y^4}}\cdot\sqrt[4]{x^2+y^4}\log(x^2+y^4).
$$
Now, it should be rather easy to see that
$$
\lim_{(x,y)\to(0,0)}\sqrt[4]{x^2+y^4}\log(x^2+y^4)=0.
$$
In fact, set $z=g(x,y)=x^2+y^4,$ prove that $g(x,y)\to0$ (use the continuity), then
$$
\lim_{z\to0}\sqrt[4]{z}\log(z)=0,
$$
while the other factor is bounded in a neighbourhood of $(0,0).$ We can equivalently consider its fourth power
$$
\frac{(x+y)^4}{x^2+y^4},\qquad x,y\geq0,\ (x,y)\neq(0,0),
$$
and use polar coordinates
$$
\frac{\rho^2(\cos\theta+\sin\theta)^4}{\cos^2\theta+\rho^2\sin^4\theta},\qquad\rho>0,\ 0\leq\theta\leq\pi/2
$$
It is easy to see that, whenever $0<\rho<1/\sqrt{2}$, the denominator satisfy
$$
\cos^2\theta+\rho^2\sin^4\theta\geq\rho^2
$$
so that
$$
0<\frac{(x+y)^4}{x^2+y^4}=\frac{\rho^2(\cos\theta+\sin\theta)^4}{\cos^2\theta+\rho^2\sin^4\theta}\leq(\cos\theta+\sin\theta)^4\leq4
$$
and
$$
0<\frac{x+y}{\sqrt[4]{x^2+y^4}}\leq\sqrt{2},\qquad x,y\geq0,\ 0<x^2+y^2<1/2.
$$
</s>