# Determine whether a function is differentiable

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?

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

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.