Let ${ f(x,y) = \frac{x}{ \left| y \right| } \sqrt{ x^{2} + y^{2} } }$ for $y \ne 0$ and $f(x,y) = 0$ for $y=0$. Does it differentiable in $(0,0)$?

I was able to prove that the function has directional derivative in any direction at $(0,0)$, but I know it doesn't mean that the function is differentiable at this point... I tried to check by differential definition at the point:

${ \lim_{(x,y)\to(x_0,y_0)} \frac{f(x,y)-f(x_0,y_0) - A \triangle x - B \triangle y}{ \sqrt{ {(\triangle x)}^{2} + {(\triangle y)}^{2}} } \iff 0}$

But I got:

${ \lim_{(x,y)\to(0,0)} \frac{f(x,y)-f(0,0) - A \triangle x - B \triangle y}{ \sqrt{ {(\triangle x)}^{2} + {(\triangle y)}^{2}} } = ... = \lim_{(x,y)\to(0,0)} = \frac{x}{|y|} }$

How should I continue from this point?

  • 2
    $\begingroup$ How is $f$ defined when $y=0$?. $\endgroup$ – Kabo Murphy Jun 14 at 9:46
  • $\begingroup$ @KaviRamaMurthy Updated the question, thanks! $\endgroup$ – Nave Tseva Jun 14 at 9:46

Your function is not continuous in zero.

Consider a path $x=t$ and $y=t^3$ for $t>0$. Your function becomes $$ \frac{t}{t^3}\sqrt {t^2+t^6} = \frac 1t \sqrt {1+t^4}. $$

Clearly, for $t\to0$ the above expression goes to $+\infty$.

On the other hand, if you consider a path $x=y=t$, your function would go to $0$ as $t\to0$, hence $f$ is not continuous (and therefore, not differentiable) in zero.


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