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Let $f: \mathbb{R}^2 \to \mathbb{R}$ be defined by $f(x)=\begin{cases} \frac{x_1^2x_2}{x_1^2+x_2^2}, \quad x\neq 0 \\ 0 , \quad \quad \, \, \, \, x=0\end {cases}$.

Show that all directional derivates in $0\in \mathbb{R}^2$ exist.

My attempt:

Let $a:=(0,0)$ and $v:=(x,y)$ with $x\neq y$, then:$$\partial _v f(0)=\lim\limits_{t\to 0}\frac{f(0+t(x,y))-f(0)}{t}=\lim\limits_{t\to 0}\frac{f(xt,yt)}{t}=\lim\limits_{t\to 0}\frac{\frac{(xt)^2(yt)}{(xt)^2+(yt)^2}}{t}=\lim\limits_{t\to 0}\frac{t^3x^2y}{t^3(x^2+y^2)}=\lim\limits_{t\to 0}\frac{x^2y}{x^2+y^2}=\frac{x^2y}{x^2+y^2}$$ But that is just valid iff $x\neq y$. Do I have to show $x=y$ seperately?

Martin Sleziak
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Analysis
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1 Answers1

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Your derivation is fine, for $x=y=c$ we obtain

$$\lim\limits_{t\to 0}\frac{x^2y}{x^2+y^2}=\frac{x^2y}{x^2+y^2}=\frac12c$$

what we need is $(x,y)\neq (0,0)$.

Note also that usually we assume $|(x,y)|=1$ and in this case we have

$$\lim\limits_{t\to 0}\frac{x^2y}{x^2+y^2}=\lim\limits_{t\to 0}x^2y$$

Refer also to the related

user
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