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I have a function $f \in c^n$ such that:

  • $f(0, 0) = 0$
  • $f(x,y) \neq 0 (\forall (x,y) \neq (0,0))$

I want to prove that the parcial derivatives at the origin are both 0. The thing is that I'm stuck with these limits:

  • $\lim_{t \to 0} \frac{f(t,0)}{t}$
  • $\lim_{t \to 0} \frac{f(0,t)}{t}$

How can I solve those indeterminations?

It is valid to state that since $f$ has continuous derivatives, both can't go to infinity and therefore those limits go to 0?

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    You want to assume the partial derivative is nonzero, then reach a contradiction. [PS: It would help to define your notation $c^n$, for example, we do not know what $n$ means.] – 2017-02-25
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    You can use the implicit/inverse function theorem. If it isn't the case $\nabla f=0$, then prove that $f$ changes signs in a neighborhood of $0$. – 2017-02-25
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    Would it work to apply the Implicit Function Theorem and state that the implicit function $\phi(x)$ is defined for only one point and then it would contradict the fact that it is differentiable? – 2017-02-25
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    Alternatively, the argument could be that the I.F.T guarantees there are **open sets**, which in this case is false since both only contain $0$. – 2017-02-25
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    @John Can you solve the one dimensional version of this problem? If you can, what keeps you from adapting the solution to $\mathbb R^2$? – 2017-02-26
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    Would the one dimensional version use the fact that $x=0$ is the only place where $y=g(x)$ can change signs? – 2017-02-26

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