# using sequences to prove discontinuity

• Mar 10th 2010, 10:01 PM
eccc2004
using sequences to prove discontinuity
Question: Prove at x=0 that cos(1/x) does not have a removable discontinuity.

Attempt: I used two sequences that as limit n-> infinity, each sequence goes to zero, but if I insert the sequences into cos(1/x), cos(1/{xn}) goes to -1 and cos(1/{yn}) goes to 1. Does this show that at x=0, cos(1/x) does not have a removable discontinuity?
• Mar 11th 2010, 03:50 AM
HallsofIvy
Yes, it does. In order to have a removable discontinuity, the limit itself must exist there. If the limit exists, then the limit along any sequence converging to that point must be equal to that limit. If two different sequences, both converging to 0, give two different results, then the limit cannot exist.
• Mar 11th 2010, 03:51 AM
tonio
Quote:

Originally Posted by eccc2004
Question: Prove at x=0 that cos(1/x) does not have a removable discontinuity.

Attempt: I used two sequences that as limit n-> infinity, each sequence goes to zero, but if I insert the sequences into cos(1/x), cos(1/{xn}) goes to -1 and cos(1/{yn}) goes to 1. Does this show that at x=0, cos(1/x) does not have a removable discontinuity?

Of course it does! it shows that $\displaystyle \lim_{x\to \infty}=cos\left(\frac{1}{x}\right)$ doesn't exist and thus the discontinuity cannot be removable.

Tonio