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Let [itex]f(x,y) = \begin{array}{cc}

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

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

\end{array}[/itex]

Show that the directional derivatives at (0,0) in directions [itex]a\mathbf{i} + b\mathbf{j}[/itex] with [itex]a\neq 0[/itex] and [itex]b\neq 0[/itex], do not exist.

Let [itex]\mathbf{u} = a\mathbf{i} + b\mathbf{j}[/itex]

[itex]

\begin{align*}

D_{\mathbf{u}}f(0,0) & = \lim_{h\rightarrow 0}\frac{f(0+ ha,0 + hb) - f(0,0)}{h}

& = \lim_{h \rightarrow 0} \frac{\frac{h^2ab}{\sqrt{h^2a^2 + h^2b^2}}}{h}

= \frac{ab}{\sqrt{a^2 + b^2}}

\end{align*}

[/itex]

which if I'm not mistaken, exists. How do I show that this doesn't exist? Also, in order to show that f(x,y) is everywhere continuous, will it suffice to say that xy/sqrt(x^2 + y^2) is continuous when (x,y) != 0 and that the limit of xy/sqrt(x^2 + y^2) as (x,y) tends to (0,0) along the x-axis is 0?

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

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

\end{array}[/itex]

Show that the directional derivatives at (0,0) in directions [itex]a\mathbf{i} + b\mathbf{j}[/itex] with [itex]a\neq 0[/itex] and [itex]b\neq 0[/itex], do not exist.

Let [itex]\mathbf{u} = a\mathbf{i} + b\mathbf{j}[/itex]

[itex]

\begin{align*}

D_{\mathbf{u}}f(0,0) & = \lim_{h\rightarrow 0}\frac{f(0+ ha,0 + hb) - f(0,0)}{h}

& = \lim_{h \rightarrow 0} \frac{\frac{h^2ab}{\sqrt{h^2a^2 + h^2b^2}}}{h}

= \frac{ab}{\sqrt{a^2 + b^2}}

\end{align*}

[/itex]

which if I'm not mistaken, exists. How do I show that this doesn't exist? Also, in order to show that f(x,y) is everywhere continuous, will it suffice to say that xy/sqrt(x^2 + y^2) is continuous when (x,y) != 0 and that the limit of xy/sqrt(x^2 + y^2) as (x,y) tends to (0,0) along the x-axis is 0?

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