I did not say that ##1/z## was continuous. I said that in complex analysis we do not have a problem with ##1/z##. It is called a pole.Office_Shredder said:@Svein I'm not sure how that addresses the question of whether ##1/z## is a continuous function though, which is the analogous question to ##1/x## in ##\mathbb{R}##
A removable discontinuity in a function occurs when there is a point on the graph where the function is undefined, but the limit of the function at that point exists. This means that the function can be made continuous at that point by simply redefining the function at that point.
A removable discontinuity can be identified by looking for a hole or gap in the graph of the function. This indicates that there is a point where the function is undefined, but the limit of the function at that point exists.
A removable discontinuity can be caused by a factor in the function that cancels out, resulting in a hole or gap in the graph. For example, in the function ##f(x) = \frac{(x-1)(x+2)}{x-1}##, the factor of (x-1) cancels out, resulting in a removable discontinuity at x = 1.
To find the limit of a function at a removable discontinuity, you can simply evaluate the function at the point where the discontinuity occurs. This will give you the value of the limit, as the function is continuous everywhere except at that point.
Yes, a removable discontinuity can be removed by redefining the function at the point where the discontinuity occurs. This can be done by filling in the hole or gap in the graph with the correct value, making the function continuous at that point.