If exists then has a pole of order . But take and then and which means does not exist ie. the function has an essential singularity at .
Let where .
Decide whether f is continous at 0 and explain your answer.
I don't think is continuous. However, I am having some trouble constructing two sequences that tend to 0 but have different limits when put into .
I think the sequences are going to be like but they aren't working!
Thanks but I managed to prove it another way!
I'm afraid my course hasn't covered what a "pole of order " is and what an essential singularity is.
I'll read up on those, thankyou!
EDIT: Those sequences are real but z is complex. I can prove that is not continous quite easily, it's just that when there's a z in front it makes things a bit trickier (this is kind of what I mean):
Similarly:
Hence they have the same limit so in is continuous. To me this implies that you would need two sequences, one approaching 0 from the imaginary axis and one approaching 0 from the real axis that have the same limit (ie. 0) but f(sequence 1) f(sequence 2) (when limits are applied). This is how I did it but it's about a page long. What you've done is shorter!