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Let be $f:\mathbb{R^2}\rightarrow\mathbb{R}$ defined by:

$$f(x,y)= \begin{cases} x^3\log{\left(1+\frac{|y|^\alpha}{x^4}\right)} & \text{if } x \neq 0 \\ 0 & \text{if } x =0 \end{cases}$$

Find for which $\alpha>0$ the function $f$ is:

  1. continuous in $\mathbb{R^2}$;
  2. differentiable in $\mathbb{R^2}$;
  3. of class $C^1$ in $\mathbb{R^2}$.

My try: I'm blocked at point 1). I tried to study the limit: $$\lim_{r\rightarrow0}\sup_{\theta}r^3\cos^3\theta\log{\left (1+ \frac{|r\sin\theta|^\alpha}{(r\cos\theta)^4} \right)}$$ but without great success. Some hints?

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First we make some computations: $$ \lim_{x\to 0, y\to y_0}x^3\log\Big(1+\frac{|y|^\alpha}{x^4}\Big)=\lim\left(x^3\sqrt{1+\frac{|y|^\alpha}{x^4}}\right)\frac{2\log(u)}{u}=\lim x\sqrt{x^4+|y|^\alpha}\cdot \frac{2\log(u)}{u}, $$ where $u=\sqrt{1+\frac{|y|^\alpha}{x^4}}$. If $(x,y)\to(0,a)$ and $u$ remains bounded, the last product goes to $0$ as the first factor dominates. But if $u\to+\infty$, we know that $\log(u)$ grows slower than $u$ and $\log(u)/u\to0$ and we are done again. This settles continuity and gives a way to treat limits involving this weird $\log$.

Next for differentiability we look at $\frac{\partial f}{\partial y}(a,0)$, $a\ne0$. Directly by definition we must compute the limit when $t\to 0$ of the quotient $$ A=\frac{f(a,t)-f(a,0)}{t}=\frac{a^3\log\Big(1+\frac{|t|^\alpha}{x^4}\Big)}{t}. $$ It's of the form $\frac{0}{0}$, hence we apply l'Hôpital: $$ \lim_{t\to0}A=\pm a^3\lim_{t\to0}\frac{\alpha|t|^{\alpha-1}}{1+\frac{|t|^\alpha}{a^4}}=\begin{cases}0&\text{for $\alpha>1$,}\\\pm\alpha a^3&\text{for $\alpha=1$,}\\ \pm\infty&\text{for $\alpha<1$.}\end{cases} $$ Here the $\pm$ distinguishes $t\to0^+$ and $t\to0^-$, and we see the limit doesn't exist for $\alpha\le1$. Consequently the function is not differentiable for $\alpha\le1$.

For the case $\alpha>1$ I summarise: clearly $f$ is ${\mathcal C}^1$ off the axis $xy=0$, so one focuses at those axis. Then compute partial derivatives off them through $x^3\log...)$ formula valid there to get $$ \begin{cases} \frac{\partial f}{\partial x}(x,y)=3x^2\log\Big(1+\frac{|y|^\alpha}{x^4}\Big)-\frac{4x^2|y|^\alpha}{x^4+|y|^\alpha},\\ \frac{\partial f}{\partial y}(x,y)=\frac{\pm \alpha x^6|y|^{\alpha-1}}{x^4+|y|^\alpha}. \end{cases} $$ On the other hand right from the definitions, one sees that all partial derivatives vanish at every point of the two axis. Next one computes the limits when $x\to0$ and when $y=0$ (two different cases) of the formulas above. As far as I've had the patience of doing it, the limits are indeed $0$, hence the function is ${\mathcal C}^1$ for $\alpha>1$.

For anotehr suggestion, let's see one limit of a partial derivative when $x\to0,\ y\to0$. WE bound as follows $$ \left|\frac{\pm \alpha x^6|y|^{\alpha-1}}{x^4+|y|^\alpha}\right|\le \alpha x^2\frac{x^4}{x^4+|y|^\alpha}|y|^\alpha, $$
and note that the frist and third factor go to $0$, and the second is bounded: since $x^4+|y|^\alpha\ge x^4$, the quotient is $\le1$.

I hope I haven't go wrong in the computations, but I think the whole lot can be of some help in any case. For other thing, I would recommend beware polar coordinates, they are useful but often lead to some confusions, see my post

Multivariable limit with polar coordinates

on the matter.

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  • $\begingroup$ Just to complete the above post. To find the condition $\alpha>1$ one can argue that when $f$ is differentiable, so is $e^{f(1,y)}-1=|y|^\alpha$, hence $\alpha>1$. But this somehow hides the insides... $\endgroup$ – Jesus RS Jan 11 '15 at 13:20

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