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Here is a problem from an old comprehensive exam that I am trying to solve

Problem: let $f:\mathbb{R}^{2} \to \mathbb{R}$ be a function defined as follows:

$f\left ( x,y \right )=\frac{\left ( x^{2}-y \right ).y^{2}}{x^{4}+y^{2}}$ if $\left ( x,y \right )\neq \left ( 0,0 \right )$

$f\left ( x,y \right )=0$ if $\left ( x,y \right )=\left ( 0,0 \right )$

The question is : Investigate the differentiability of $f$ at the point $\left ( 0,0 \right )$

Here is what I did so far: $\frac{\partial f}{\partial x}\left ( 0,0 \right )=0 $ and $\frac{\partial f}{\partial y}\left ( 0,0 \right )=-1 $

Now I applied the condition for differentiability for $f$ at the point $\left ( 0,0 \right )$:

$lim_{\left ( x,y \right ) \to \left ( 0,0 \right )}\frac{\left \| f\left ( x,y \right )-f\left ( 0,0 \right )-\bigtriangledown f\left ( 0,0 \right ).\left ( x,y \right ) \right \|}{\sqrt{x^{2}+y^{2}}}$ has to be $0$ if $f$ is differentiable at the desired point.

After simplifying the above limit, I got the following limit:

$lim_{\left ( x,y \right ) \to \left ( 0,0 \right ) }\frac{x^{2}y^{2}+x^{4}y}{\left ( x^{2}+y^{4} \right ).\sqrt{x^{2}+y^{2}}}$ . Here is where I am stuck. I cannot evaluate this limit. I tried everything, like evaluating the limit through different paths... but nothing seems to work out for me.

Any help please on how to finish my proof?

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up vote 1 down vote accepted

$\bf Hint:$ Using polar coordinates the equation turns into: $lim_{r\to 0}\frac{r^4cos^2\theta sin^2\theta+r^5cos^4\theta sin\theta}{r^3cos^2\theta+r^5sin^4\theta} $

Edit: Note that if $sin\theta=0$ then the equation turns into $lim_{r\to 0}\frac{r^4+r^5}{r^3}=lim_{r\to 0}r+r^2=0$.

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Ok, using polar coordinates, I got the final result: $lim_{r\rightarrow 0^{+}}\frac{r.cos^{2}\theta.sin^{2}\theta +r^{2}cos^{4}\theta sin\theta }{r^{2}cos^{4}\theta +sin^{2}\theta }$. But, as $r\rightarrow 0^{+}$, we get: $\frac{0}{sin^{2}\theta }$. Does this mean that the limit is zero? But we have no idea what $sin^{2}\theta$ can be. It can be zero if $\theta$ is zero, and we get an indeterminate form of $\frac{0}{0}$ – M.Krov Feb 4 '12 at 4:28
One more question. Suppose the limit turned out to be dependent on $\theta$ like $sin(\theta)$. Does this mean that the limit doesn't exit and the function is not differentiable? – M.Krov Feb 4 '12 at 4:33
I edited for the case $sin \theta=0$. If the answer depends on $\theta$ the limits depends on the angle you are approaching zero. – azarel Feb 4 '12 at 4:42
Just to comment on your answer: If $sin(\theta)$ is $0$. The numerator will vanish and equal to $0$ and the denominator should be $r^{2}$. Am I correct? OR is there anything I missed because I can't get your result? Also, would you please answer my question above? Thanks – M.Krov Feb 4 '12 at 4:44
Ok, then if the limit depends on the angle of approach $\theta$, then does this mean that the function is not differentiable? – M.Krov Feb 4 '12 at 4:46

You can separate the limit (that you got in the definition of diferentiability) as the limit of a sum. Each sumand : you can express as the products of bounded functions multiplied by functions that converge to zero: X^2/x^2+y^4 is bounded, y/square root of x^2+y^2 is bounded. You know that a bounded function multiplied by a factor that converges to 0 , gives a product that converges to zero.

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