# Formal definition on differentiability - an excerise

Let $f: \mathbb{R}^2 \to \mathbb{R}$ be defined by setting $f(\mathbf{0}) = 0$ and $$f(x,y) = xy/(x^2 + y^2)$$ if $(x,y) \neq (0,0)$

(a) For which vectors $\mathbf{u} \neq \mathbf{0}$ does $f'(\mathbf{0}; \mathbf{u})$ exist? Evaluate it when it exists.

(b) Is $f$ differentiable at $\mathbf{0}$? Is it continuous at $\mathbf{0}$?

Okay I am a little bothered by my answer for the first question because my computation yeilds a limit which doesn't exist and from the question it seems to imply the directional derivative exists.

(a) $\lim_{t\to 0} \dfrac{f(0+t\mathbf{u}) - f(\mathbf{0})}{t} = \lim_{t \to 0} \dfrac{f(t\mathbf{u})}{t} = \lim_{t\to0} \dfrac{tu_1u_2}{t^2 (u_1^2 + u_2^2)} = \lim_{t\to0} \dfrac{u_1u_2}{t (u_1^2 + u_2^2)}$ which doesn't exist.

(b) For this one, I also got a limit which doesn't exist.

$\lim_{\mathbf{h} \to\mathbf{0}} \dfrac{f(\mathbf{0}+\mathbf{h})- f(\mathbf{0}) - B\cdot \mathbf{h}}{|\mathbf{h}|} = \lim_{\mathbf{h} \to\mathbf{0}} \dfrac{f(\mathbf{h})- B\cdot \mathbf{h}}{|\mathbf{h}|} = \lim_{(h_1,h_2) \to(0,0)} \dfrac{h_1 h_2 - b_1 h_1 - b_2 h_2}{(h_1^2+h_2^2 )\sqrt{h_1^2 + h_2^2}} =\lim_{(h_1,h_2) \to(0,0)} \dfrac{h_1 h_2 - b_1 h_1 - b_2 h_2}{(h_1^2+h_2^2 )^{3/2}}$ which also does not exist, so it can't be continuous at 0

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Isn't the same thing? I factored it out no? –  Hawk Jan 25 '13 at 23:08
Sorry, you skipped a step and I missed it. –  André Nicolas Jan 25 '13 at 23:11

OK, so for (a), your solution is almost right. Notice that the limit does exist if $u_1$ or $u_2$ are $0$, in which case the directional derivative does exist.
In (b) note that $f$ is not continuous at $0$ since approaching $(0,0)$ along the curve $(0,t)$ yields the limit $0$, while approaching it along $(t,t)$ yields the limit $1/2$. If $f$ is not continuous at $0$ it cannot be differentiable.
How do I show that limit does not exist? I managed to substitute $h_1 = mh_2$ and reduced the stuff inside the limit to $$\dfrac{my - (am + b)}{y^2 (m^2 + 1)^(3/2)}$$ –  Hawk Jan 25 '13 at 23:56
You could probably get something doing that, but it's simpler than that. If the function is differentiable, then it must be continuous, but the function is not continuous (using the two paths I wrote in the previous comment), therefore $f$ cannot be continuous. –  William Jan 26 '13 at 0:15
Can I comment that from (a) that it is differentiability only in the direction of $i$ and $j$? –  Hawk Jan 26 '13 at 0:18