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Let $f : \mathbb{R}^2 \to \mathbb{R}$ be defined by $$f(x,y) = \begin{cases} \frac{x^2 y^2}{x^4 + y^4}, & \text{ }\text{(x,y)} \neq (0,0) \\ 0, & \text{ }\text{(x,y)} = (0,0) \end{cases} .$$

Show that $\frac{df}{dx} (0,0)$ exists, and $f$ is not continuous at $(0,0)$.

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Write down the limit defining the partial derivative. What do you get? Compute the limit of $f$ along the path $x=y$ and compare with that value of $f(0,0)$. What do you get? – David Mitra Apr 2 '12 at 23:27
up vote 2 down vote accepted

For existence of the partial derivative:

By definition, $f_x(0,0)$ is the limit $$ \lim_{h\rightarrow 0}{f(h,0)-f(0,0)\over h }. $$ Show that this limit exists for your particular function.

For continuity:

Note that,
$$ \lim_{x\rightarrow 0} f(x,x) =\lim_{x\rightarrow 0}{x^4\over2x^4}={1\over2}. $$ Keeping this in mind and the fact that $f(0,0)=0$, what can you say about the continuity of $f$ at $(0,0)$?

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