Does a tangent exist at $x=0$ to $y=sgn(x)$?

Question

Yesterday my professor told me that a tangent can be constructed at $x=0$ to the signum function reasoning that the two points considered while drawing a tangent must be close horizontally and not necessarily vertically.

He also told me that existence of tangent(s) isn't sufficient to prove that a derivative exists, i.e. a tangent can be drawn even when there exists no derivative.

Can anyone elaborate this point? Also, regarding the first statement, will this: $x=sgn(y)$ have a tangent at $x=0$? In this case, the two points now are close vertically and not horizontally...

Thanks in advance.

Answer 1

"Does a tangent exist at x=0 to y=sgn(x)?."

It all depends on how you are defining the tangent. If you are calling the line with the slope equal to the derivative at a point as a tangent, then the answer would be no, because now sgn(x) is discontinous.

If you are calling the limiting chord a tangent then the Y-axis is tangent at x=0.

Answer 2

yes a vertical tangent can be constructed at x=0. use first derivative principle to calculate left hand derivative and right hand derivative at x=0 , both will come infinity.Hence a vertical tangent .