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Given the function $$f(x,y)=\begin{cases} \sqrt{x^2+y^2} . \sin\left (\frac{y^2}{x} \right ) & \text{ if } x \neq 0 \\ 0 & \text{ if } x= 0 \end{cases}$$

I have to check the continuity and differentiability of function at $(0,0)$

Solution I tried I used the sufficient condition for continuity

A sufficient condition that a function f be continuous at (a,b) is that one of the partial derivatives exists and is bounded in a neighbourhood of (a,b) and other exists at that point

so i find the $f_x$ and $f_y$ at point $(0,0)$ and they both are zero ,so function is continuous at $(0,0)$

for differentiability i tried $$\lim_{(x,y) \to (0,0)}\frac{||f(x,y)-xf_x(0,0)-yf_y(0,0)-f(0,0)||}{||(x,y)||}$$

i got $$\lim_{(x,y) \to (0,0)}\frac{\sqrt{x^2+y^2} || \sin\left (\frac{y^2}{x} \right ) ||}{\sqrt{x^2+y^2}}$$

$$\lim_{(x,y) \to (0,0)} || \sin\left (\frac{y^2}{x} \right ) ||$$

next if i take polar cordinate $x=r\cos \theta$ and $y=r\sin \theta$

$$\lim_{r \to 0} || \sin\left (\frac{r \sin^2 \theta}{ \cos \theta} \right ) || \to 0$$

it goes to zero , so the function is differentiable at $(0,0)$

but i am not confirmed about last calculation , please help me to check this solution

Thank you

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The statement

$$\lim\limits_{r \to 0} \sin \left(r\frac{\sin^2 \theta}{\cos \theta}\right)=0$$ is correct for $\theta$ fixed.

Unfortunately, this is not sufficient to evaluate the limit at the origin.

As you noticed, if the derivative would exist at the origin, it would be the always vanishing linear map.

But $$\lim\limits_{y \to 0} \frac{f(y^2, y)}{\sqrt{y^4+y^2}}= \sin 1 \neq 0$$ proving that $f$ is not differentiable at the origin.

The continuity at the origin is clear as

$$\vert f(x,y)\vert \le \sqrt{x^2+y^2}$$ for all $(x,y) \in \mathbb R^2$.

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  • $\begingroup$ I am not getting the line "if the derivative would exist at the origin, it would be the always vanishing linear map" can you please elaborate this? $\endgroup$
    – TheStudent
    Jun 17 at 4:37
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    $\begingroup$ This is the consequence of the fact that both partial derivatives vanish at the origin. $\endgroup$ Jun 17 at 7:54

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