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Let $f $ be the function on $R^2$ defined by $f (x, y) = \dfrac {x^3}{y}$, if $y = 0$,

$0$, if $y = 0$.

(i) Prove that the directional derivative $D_vf (0, 0)$ (exists and) $= 0$ for each $v ∈ R^2$.

(ii) $f $ is not continuous at $(0, 0)$.

(iii) $f $ is not differentiable at $(0, 0)$.

I'm not able to prove item (ii), in fact i'm getting the opposite result: f is continuous at $(0,0)$

I thought the $lim_{(x,y)\rightarrow (0,0)} f (x) = lim_{(x,y)\rightarrow (0,0)} \dfrac {x^3}{y}$ and i thought this goes to 0, because $x^3$ goes faster to 0 then $y $. Is that right?

I also tried to do it by using the straight definition of the limit but i could'nt manage to give that proof

AnaliseMat
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  • Could you tell us how you're proving that this is continuous at $(0,0)$? We really can't help you with resolving that if you don't show us the work you've done. – Milo Brandt Sep 17 '19 at 00:48
  • You know $f(0,0)=0$. Check that $f(1/n,1/n^3) \not \to 0$ even though $ (1/n,1/n^3) \to (0,0)$ – David P Sep 17 '19 at 00:49
  • Milo Brandt, sorry, i atualized my post with some work that i have done – AnaliseMat Sep 17 '19 at 00:56

1 Answers1

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to show that it is not continuous, take $y=x$ as they approach $(0,0)$. This gives

$\frac{x^3}{y}=x^2=0$. Then take the path $y=x^3$. This gives $\frac{x^3}{y}=\frac{x^3}{x^3}=1=1$. You can also take other paths such as $y=x^4$ to get $\frac{1}{x}$ as $x$ goes to zero. Which is clearly not zero.

Sina Babaei Zadeh
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