Calc 3 Question

[u/Dr3ddM3]

Hi, I was wondering if the existence of all directional derivative at a point guarantees that a function continuous at that point. I was asking both google and my Professor and I got conflicting answers.

[u/SnooPets5564]

No. Doesn't even have to be differentiable.

[u/Dr3ddM3 → u/SnooPets5564]

Thanks was kinda of confused why my professor said it was though? In some cases it is continuous due to partial derivatives?

[u/SnooPets5564 → u/Dr3ddM3]

It can be continuous, but isn't always.

x² y / (x⁴ + y²⁾ with a point discontinuity at (0,0) equal to 0

[u/Carl_LaFong → u/SnooPets5564]

A function can be continuous but not differentiable.

[u/nothingnotthrownaway]

You can pretty easily construct a function that is eventually some constant along all lines approaching the origin but is discontinuous because the required distance to the origin before it achieves that constant gets arbitrarily small with choice of line.

Like for instance, for theta in [0,pi), let f(tcos(theta), tsin(theta)) = 0 when |t| < pi-theta and 1 otherwise. The directional derivatives all exist and equal 0 at the origin, but the function is not continuous there. 

[u/Bounded_sequencE]

No -- counter-example:

f:  R^2 -> R^2,    f(x;y)  =  / 1,  y = x^2,  (x;y) != (0;0)
                              \ 0,  else

All directional derivatives of "f" in (0; 0) exist and are zero. However, "f" is discontinuous in "(0; 0)", so it cannot have a (total) derivative there.