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$f:\mathbb{R}^2\rightarrow \mathbb{R}$ Defined by $$f(x,y)= \frac{xy^2}{x^2+y^4}$$ if $x\neq 0,y\in\mathbb{R}$ and $$f(x,y)=0$$ if $x=0,y\in\mathbb{R}$

Then

  1. it is continuous but not differentiable at origin

  2. differentiable at origin

  3. has all first order partial derivative at origin.

  4. does not have all first order derivatives at origin.

consider the limit $(x,y)\rightarrow (0,0)$ along the curve $y=m\sqrt{x}$ we get lim$$(x,y)\rightarrow(0,0)\frac{x^2m^2}{x^2+m^4x^2}=\frac{m^2}{1+m^4}$$ which is different for different values for $m$ hence $f$ is not continuous at origin, so 1 is false, and 2 is clearly false. I have checked that $f_x$ and $f_y$ exists at $(0,0)$ so only $3$ is correct and all others are false. could any one confirm me am I right? Thank you.

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    The question "What more can I say" has infinitely many distinct answers, I suppose :-) Can you make it a bit more precise?2012-07-12
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    @Siminore edited :)2012-07-12
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    I think the titled should be more telling.2012-07-12
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    By $f(x)$ do you mean $f(x,y)$?2012-07-12
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    edited @Mercy thank you2012-07-12
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    I think you'd better add more information into the title. It will make the question potent. :-)2012-07-12

1 Answers 1

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$\displaystyle \lim_{x\rightarrow0}\frac{f(x,0)-f(0,0)}x$is found to be 0. Similarly for $y$. So 3 is true and not 4

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    thank you aneesh got it. :)2012-07-12