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Let $f:\mathbb R^n\longrightarrow \mathbb R$ and $a\in\mathbb R^n$ s.t. $\nabla f(a)$ exist.

I know that for a function $g:\mathbb R\longrightarrow \mathbb R$, it suffice that $g'(a)$ exist to have that $g$ is differentiable in $a$. Indeed, $$\lim_{h\to 0}\frac{g(a+h)-g(a)-g'(a)h}{h}=\lim_{h\to 0}\left(\frac{g(a+h)-g(a)}{h}-g'(a)\right)=0$$ since $$\lim_{h\to 0}\frac{g(a+h)-g(a)}{h}=g'(a).$$ Now I was wondering why in $\mathbb R^n$ the continuity of the gradient is requiert. So I try to compute :

$$\lim_{h\to 0}\frac{f(a+h)-f(a)-\nabla f(a)\cdot h}{\|h\|},$$ but I don't see in what the fact that $\nabla f$ is continuous at $a$ is necessary to conclude that the limit above is $0$. I mean, when we do the calculation, when do we need the fact that $\nabla f$ is continuous at $a$ ?

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    Continuity of $\nabla f$ at $a$ is not required. – 2017-02-12

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I think you have some confusion about what the theorems say. I don't have the proofs or the counter examples available to me right now. But here is at least something for you.

The definition of the derivative of a function $f:R^2 \to R$ is as follows. At $a$, a derivative of $f$ exists at $a$ if there is a linear function $L: R^2 \to R$ such that $$\lim_{h \to 0} \frac{f(a+h)-f(a)-L(h)}{||h||} = 0.$$ Remember that $h$ is a vector. We call $L$ the derivative.

This is a stronger condition than the partial derivatives of $f$ existing. That is, for $f(x,y)$, both $$\lim_{\delta x \to 0} \frac{f(y,x+\delta x)-f(x,y)}{\delta x}$$ and $$\lim_{\delta y \to 0} \frac{f(y+\delta y,x)-f(x,y)}{\delta y}$$ existing does not necessarily imply that there is a linear map $L$ that satisfies the above. If you want there to be just a linear map $L$, then you need additional conditions. Namely, that both $f_x$ and $f_y$ are continuous. I don't have the proof available to me so I forget the precise reason why. There is a proof of this in Rudin's Principles of Analysis.

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    To give an explicit counterexample, $f(x,y) = xy/\sqrt{x^2+y^2}$ is continuous and its partial derivatives exist everywhere, but the limit in the question does not exist at $(0,0)$. – 2017-02-12
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The continuity of $\nabla f$ at $a $ is not necessary !

If all partial derivatives of $f$ at $a$ exist, then we have:

f is differentiable at $a$ $\iff$

$\lim_{h\to 0}\frac{f(a+h)-f(a)-\nabla f(a)\cdot h}{\|h\|}=0$

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    I have a theorem that says that $f$ is differentiable in $a$ $\iff$ all partial derivative exists and are continuous in $a$, which is equivalent to $\nabla f$ continuous in $a$. – 2017-02-12
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    @MSE but your claim doesn't hold in 1d. – 2017-02-12
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    Why downvote ? ?? – 2017-02-12
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    @MSE There is no such theorem. – 2017-02-12
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    @LovroSindičić I do not understand your comment. – 2017-02-12
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    @Fred I have no idea why people are down voting your answer. It is completely correct. I upvoted it, and anyone who down voted it should do the same. – 2017-02-12