Results 1 to 3 of 3

Math Help - continuity in C

  1. #1
    Super Member Showcase_22's Avatar
    Joined
    Sep 2006
    From
    The raggedy edge.
    Posts
    782

    continuity in C

    Let f(z)= \begin{cases} z \cos \left( \frac{1}{z} \right) \ \ \ x \neq 0 \\<br />
0 \ \ \ \ \ \ \ \ \ \ \ \ \ x=0<br />
\end{cases} where z \in \mathbb{C}.

    Decide whether f is continous at 0 and explain your answer.

    I don't think f(z) is continuous. However, I am having some trouble constructing two sequences that tend to 0 but have different limits when put into f(z).

    I think the sequences are going to be like a_n=\frac{1}{2n \pi+ \frac{\pi}{2}} but they aren't working!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Apr 2009
    From
    México
    Posts
    721
    If \lim_{z\rightarrow 0} z^k \cos ( \frac{1}{z} ) exists then \cos (\frac{1}{z} ) has a pole of order \leq k. But take a_k= \frac{1}{\pi /2+ \pi k} and b_k= \frac{1}{2\pi k} then \cos (\frac{1}{a_k} )=0 and \cos (\frac{1}{b_k} )=1 which means \lim_{z\rightarrow 0} \cos (\frac{1}{z} ) does not exist ie. the function has an essential singularity at z=0.
    Last edited by Jose27; November 18th 2009 at 12:49 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member Showcase_22's Avatar
    Joined
    Sep 2006
    From
    The raggedy edge.
    Posts
    782
    Thanks but I managed to prove it another way!

    I'm afraid my course hasn't covered what a "pole of order \leq k" 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  \cos \left( \frac{1}{z} \right) 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):

     f(b_k)=b_k \cos \left( \frac{1}{b_k} \right)=0

    Similarly:

     f(a_k)=a_k \cos \left( \frac{1}{a_k} \right)=0

    Hence they have the same limit so in  \mathbb{R}, \ f(z) 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)  \neq 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!
    Last edited by Showcase_22; November 19th 2009 at 01:38 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. continuity
    Posted in the Calculus Forum
    Replies: 5
    Last Post: April 20th 2011, 01:36 PM
  2. y=x^3 (continuity)
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: May 16th 2010, 10:50 PM
  3. continuity
    Posted in the Differential Geometry Forum
    Replies: 5
    Last Post: April 9th 2010, 04:23 PM
  4. Continuity
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 7th 2009, 09:10 PM
  5. Continuity, Uniform Continuity
    Posted in the Calculus Forum
    Replies: 0
    Last Post: February 1st 2009, 08:36 PM

Search Tags


/mathhelpforum @mathhelpforum