- #1

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## Homework Statement

I need to see if the function defined as

##f(x,y) = \left\{

\begin{array}{lr}

\frac{xy^2}{x^2 + y^2} & (x,y)\neq{}(0,0)\\

0 & (x,y)=(0,0)

\end{array}

\right.##

is differentiable at (0,0)

## Homework Equations

[/B]

A function is differentiable at a point, if It can be approximated at that point by a linear transformation,

##\lim_{(u, v) \rightarrow (0, 0)} \frac {f(u, v) - df_{(0, 0)}(u, v) - f(0, 0)} {||(u, v)||} =0##

Useful bounds:

##|u|\leq ||(u, v)||;

|v|\leq ||(u, v)||##

##|u|^2=u^2\leq ||(u, v)||^2= (\sqrt{u^2+v^2})^2 = (u^2+v^2)##

## The Attempt at a Solution

[/B]

Both partial derivatives are zero at the point, so all I have left is

##\lim_{(u, v) \rightarrow (0, 0)} \frac {\frac{uv^2}{u^2 + v^2}} {||(u, v)||} =\lim_{(u, v) \rightarrow (0, 0)} \frac {uv^2}{(u^2 + v^2)||(u, v)||} ##

Now the bounds:

##\frac {uv^2}{(u^2 + v^2)||(u, v)||}\leq\frac {u||(u, v)||^2}{(u^2 + v^2)||(u, v)||} = \frac {u||(u, v)||}{(u^2 + v^2)} \leq \frac {||(u, v)||^2}{(u^2 + v^2)}= \frac{||(u, v)||^2}{||(u, v)||^2}=1##

Does this means that the function is not differentiable at (0,0), or did I make a mistake along the way ? Thanks in advance for the help.