Results 1 to 5 of 5

Math Help - Prove f(x)=x2^x is continuous on (0,1)

  1. #1
    Senior Member Pinkk's Avatar
    Joined
    Mar 2009
    From
    Uptown Manhattan, NY, USA
    Posts
    419

    Prove f(x)=x2^x is continuous on (0,1)

    I'm blanking out, how would I show either a sequential limit or epsilon-delta proof for this? For the sequential limit definition, is it an obvious property that if \lim_{n\to \infty} x_{n} = x, then \lim_{n\to \infty} 2^{x_{n}} = 2^{x}. If not, how do I go about proving that, or how would I show an epsilon-delta proof? I am completely stuck on how it could be shown using epsilon-delta. Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by Pinkk View Post
    I'm blanking out, how would I show either a sequential limit or epsilon-delta proof for this? For the sequential limit definition, is it an obvious property that if \lim_{n\to \infty} x_{n} = x, then \lim_{n\to \infty} 2^{x_{n}} = 2^{x}. If not, how do I go about proving that, or how would I show an epsilon-delta proof? I am completely stuck on how it could be shown using epsilon-delta. Thanks.
    ....isn't the multiplication of two continuous functions continuous. Or do you mean 2^x-2^y? First prove that f is continuous at zero and then note that |2^x-2^y|=|2^y||2^{\frac{x}{y}}-1|
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member Pinkk's Avatar
    Joined
    Mar 2009
    From
    Uptown Manhattan, NY, USA
    Posts
    419
    I'm sorry, I don't see how that helps. Are you trying to use that for an epsilon delta proof that f(x)=2^{x} is continuous on (0,1)? Because then I want to show that there exists delta such that |x-x_{0}|<\delta \implies |2^{x} - 2^{x_{0}}|< \epsilon for \epsilon > 0, x\in (0,1)

    Err, nevermind, I caught your error and with the correction I understand it.


    Okay, so I worked out if \delta = log_{2}(\frac{\epsilon}{2^{x_{0}}} +1), then with manipulation we get if |x-x_{0}| <\delta then |2^{x} - 2^{x_{0}}|< \epsilon

    I think that's correct, right?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by Pinkk View Post
    I'm sorry, I don't see how that helps. Are you trying to use that for an epsilon delta proof that f(x)=2^{x} is continuous on (0,1)? Because then I want to show that there exists delta such that |x-x_{0}|<\delta \implies |2^{x} - 2^{x_{0}}|< \epsilon for \epsilon > 0, x\in (0,1)
    Forget (0,1).

    Theorem: Let f:\mathbb{R}\to\mathbb{R} be such that f is continuous at 0 and f(x)f(y)=f(x+y) then f is continuous on \mathbb{R}.

    Proof:

    If f(0)=0 we're done since f(x)=f(x+0)=f(x)f(0)=f(x)0=0 and f(x)=0 is trivially continuous. Thus, assume that f(0)\ne 0 then f(0)=f(0+0)=f(0)f(0)\implies f(0)=1. Also, notice that f(x)\ne 0 for any x\in\mathbb{R} since we covered the case above assume that x\ne 0 then f(x)=0\implies f(x)f(-x)=0\implies f(0)=0 which is obviously a contradiction. It follows that f(x)f(-x)=f(0)=1\implies f(-x)=\frac{1}{f(x)} and so \frac{f(x)}{f(y)}=f(x-y)

    Thus, since f is continuous at 0 for every \varepsilon>0 there exists some \delta>0 such that |x|<\delta\implies |f(x)-1|<\varepsilon.

    So, let \varepsilon>0 be given and x_0\in\mathbb{R} arbitrary. Then, |f(x_0)-f(x)|=|f(x_0)|\left|1-\frac{f(x)}{f(x_0)}\right|=|f(x_0)|\left|1-f(x-x_0)\right|

    Now, let z=x-x_0. We know that there exists a \delta>0 such that |z|<\delta\implies |1-f(z)|<\frac{\varepsilon}{|f(x_0)|}. Thus, if |x-x_0|=|z|<\delta\implies |f(x_0)||1-f(x-x_0)|=|f(x_0)||1-f(z)|<|f(x_0)|\frac{\varepsilon}{|f(x_0)|}=\vareps  ilon. It follows that f(x) is continuous at x_0.

    Since x_0 was arbitrary it follows that f(x) is continuous on \mathbb{R}. \blacksquare.

    All you have to do is prove that f(x)=2^x is continuous at 0
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    It might be interesting to point out that while I showed the proof I did because it is interesting it turns out that the only continuous function \varphi:\mathbb{R}\to\mathbb{R} such that \varphi(x)\varphi(y)=\varphi(x+y) (and thus the only function of this form which is continuous at 0 as I prove above) is \varphi(x)=\varphi(1)^x.

    To see this first note that by induction we have that for every n\in\mathbb{N} that \varphi(n)=\varphi(\underbrace{1+\cdots+1}_{n\text  { times}})=\underbrace{\varphi(1)\cdots\varphi(1)}_{  n\text{ times}}=\varphi(1)^n. Therefore, if n\in\mathbb{N} we have that \varphi(1)=\varphi\left(\frac{n}{n}\right)=\varphi  \left(\frac{1}{n}\right)^n\implies \varphi^{\frac{1}{n}}=\varphi\left(\frac{1}{n}\rig  ht). THus, we have that if \frac{p}{q}\in\mathbb{Q} that \varphi\left(\frac{p}{q}\right)=\varphi\left(\frac  {1}{q}\right)^p=\varphi(1)^{\frac{p}{q}}.

    It follows that if we define A\left(\varphi(x),\varphi(1)^x\right)=\left\{y\in\  mathbb{R}:\varphi(y)=\varphi(1)^y\right\} we have that A(\varphi(x),\varphi(1)^x) is dense in \mathbb{R} and since it's closed it follows that A(\varphi(x),\varphi(1)^x)=\overline{A(\varphi(x),  \varphi(1)^x)}=<br />
\mathbb{R} from where the conclusion follows.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. prove that every continuous function is separately continuous
    Posted in the Differential Geometry Forum
    Replies: 6
    Last Post: November 23rd 2011, 04:57 AM
  2. Prove that f(x) is continuous.
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 3rd 2011, 02:35 PM
  3. Can anyone prove sin(x) continuous
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: May 19th 2010, 08:07 AM
  4. Replies: 6
    Last Post: November 17th 2009, 06:13 AM
  5. Prove that h is not continuous at 0
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 20th 2009, 04:44 PM

Search Tags


/mathhelpforum @mathhelpforum