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This might be a trivial question.

Consider a function $f:\mathbb{R^2}\rightarrow \mathbb{R}$ and consider some point $(a,b)\in \mathbb{R^2}$.

Suppose we know that all the directional derivatives $D_{\overline{u}}f(a,b)$ for an arbitrary unit vector $\overline{u}=\langle u_1,u_2\rangle$ in $\mathbb{R^2}$ exist $...(1)$

Furthermore, we also know that $\lim_{\overline{u}\to\overline{v}}D_{\overline{u}}f(a,b) = D_{\overline{v}}f(a,b)$ for some other arbitrary arbitrary unit vector $\overline{v}$$...(2)$

So all the directional derivatives can smoothly transition into one another as we change the unit vector $\overline{u}$ in $\mathbb{R^2}$$...(*)$

This is however $\textbf{not}$ sufficient for the function $f$ to be differentiable at $(a,b)$ since all the directional derivatives must be parallel to one another to trace out the tangent plane at the point $(a,b)$ as we change $\overline{u}$$...(*)$

$(*)$ As far as my understanding goes.

$\textbf{My Question:}$

From $(1)$ & $(2)$ can we conclude however that $f$ is continuous at $(a,b)$ ?

The only other ways of establishing continuity at $(a,b)$ in calculus of $2$ variables - that I am aware of - is by applying the$\ $ $\epsilon -\delta$ definition of continuity to $f$ at $(a,b)$ or establishing the differentiability of $f$ at $(a,b)$ which implies continuity.

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The directional derivatives are linear combinations of partial derivatives:

$$D_\overline{u}f(a,b) = u_1D_1f(a,b) + u_2D_2f(a,b),$$

so continuity with respect to $\overline{u}$ is guaranteed and not particularly relevant in terms of the continuity of $f$.

If all partial derivatives exist in an open neighborhood of $(a,b)$ and, in addition, they are continuous at $(a,b)$, then $f$ is differentiable (and continuous) there. If any partial derivative fails to exist throughout every open neighborhood or is discontinuous at $(a,b)$ then it is not guaranteed that $f$ is continuous there.

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  • $\begingroup$ Thank you for the response. I agree that is another way of establishing differentiability at a point (which then implies continuity). But, I'm struggling to see how that relates to what I'm asking in regards to $(1)$ & $(2)$. $\endgroup$ – Walt van Amstel Mar 17 '16 at 8:11
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    $\begingroup$ My point is (2) adds nothing here. It is always the case if all directional derivatives and hence partial derivatives exist. (1) and (2) do not guarantee that $f$ is continuous at $(a,b)$. $\endgroup$ – RRL Mar 17 '16 at 8:13
  • $\begingroup$ Any linear function of $\overline{u}$ is continuous. $\endgroup$ – RRL Mar 17 '16 at 8:15

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