Results 1 to 3 of 3
Like Tree1Thanks
  • 1 Post By Plato

Thread: Proving a function is continuous

  1. #1
    Junior Member
    Joined
    Sep 2009
    Posts
    62

    Proving a function is continuous

    Problem: Prove using epilson delta definition that $\displaystyle f(x)=kx$ is continuous for any real number $\displaystyle k$.

    I have the $\displaystyle k \neq 0$ case. I just had some problems with the $\displaystyle k=0$ case. Here's what I got.

    Let $\displaystyle \epsilon > 0$ and choose $\displaystyle \delta=\epsilon$, then $\displaystyle |x-x_{0}| < \delta $ implies that $\displaystyle |x-x_{0}| < \epsilon $ and so $\displaystyle |k||x-x_{0}| < |k|\epsilon $. Then $\displaystyle |kx-kx_{0}| < |k|\epsilon $ and since $\displaystyle k=0$ and $\displaystyle f(x)=kx=0$ and $\displaystyle f(x_{0})=kx_{0}=0$ then $\displaystyle |f(x)-f(x_{0})| < |k|\epsilon $. Since $\displaystyle \epsilon > 0$, then $\displaystyle \epsilon > |k|\epsilon$. So $\displaystyle |f(x)-f(x_{0})| < \epsilon$.

    Thoughts? Corrections?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,742
    Thanks
    2814
    Awards
    1

    Re: Proving a function is continuous

    Quote Originally Posted by JSB1917 View Post
    Problem: Prove using epilson delta definition that $\displaystyle f(x)=kx$ is continuous for any real number $\displaystyle k$.

    I have the $\displaystyle k \neq 0$ case. I just had some problems with the $\displaystyle k=0$ case.

    Oh, come on. If $\displaystyle k=0$ then $\displaystyle f(x)=0$ everywhere.

    Thus $\displaystyle |x-a|<\delta$ implies $\displaystyle |0-0|<\epsilon$ everywhere.
    Thanks from topsquark
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Sep 2012
    From
    Planet Earth
    Posts
    217
    Thanks
    56

    Re: Proving a function is continuous

    Quote Originally Posted by Plato View Post
    Oh, come on. If $\displaystyle k=0$ then $\displaystyle f(x)=0$ everywhere.

    Thus $\displaystyle |x-a|<\delta$ implies $\displaystyle |0-0|<\epsilon$ everywhere.
    Lol, this is the problem with drilling too much rigor into people. They start losing their sense of intuition.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. proving a function is continuous with a given topology
    Posted in the Differential Geometry Forum
    Replies: 5
    Last Post: Jan 12th 2011, 05:48 PM
  2. Proving the Continuous Function
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Dec 28th 2010, 02:24 AM
  3. Proving a function is uniformly continuous on R
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: Nov 10th 2010, 07:45 AM
  4. Proving a Function is Continuous
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: Oct 15th 2010, 03:08 AM
  5. Proving a function is continuous
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Nov 18th 2009, 08:25 PM

Search Tags


/mathhelpforum @mathhelpforum