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Math Help - Continuous functions

  1. #1
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    Continuous functions

    1) Consider the formula f(x)=lim(n-->infinity)((x^n)/(1+x^n)).
    Let D={x:f(x) is an element of R}. Calculate f(x) for all x elements of D and determine where f: D-->R is continuous.

    2) Let f: D-->R and suppose that f(x) greater than equal 0 for all x elements of D. Define sqrt(f)-->R by (sqrt(f))(x) = sqrt(f(x)). If f is continuous at c elements of D, prove that sqrt(f) is continuous at c.
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    Quote Originally Posted by noles2188 View Post
    1) Consider the formula f(x)=lim(n-->infinity)((x^n)/(1+x^n)).
    Let D={x:f(x) is an element of R}. Calculate f(x) for all x elements of D and determine where f: D-->R is continuous.

    2) Let f: D-->R and suppose that f(x) greater than equal 0 for all x elements of D. Define sqrt(f)-->R by (sqrt(f))(x) = sqrt(f(x)). If f is continuous at c elements of D, prove that sqrt(f) is continuous at c.
    We have :f(x)= \lim_{n\rightarrow\infty}{\frac{x^n}{1+x^n}} =\frac{\lim_{n\rightarrow\infty}{x^n}}{1+\lim_{n\r  ightarrow\infty}{x^n}}.................................................. .........................................1

    AND now we have the following cases:

    1) |x|<1 <===> -1<x<1 ,then \lim_{n\rightarrow\infty}{x^n} =0.

    2) x=1 ,then \lim_{n\rightarrow\infty}{x^n} =1.


    3) x>1 ,then \lim_{n\rightarrow\infty}{x^n} =\infty.


    4) x =-1 ,then x^n oscillates finitely.

    5) x<-1 ,then  x^n oscillates infinitely.

    And thus by using (1)

    For -1<x<1 ,f(x) = 0 and

    For x=1 , f(x)=1/2.

    Thus D= (-1,1] = (-1,1)U{1}

    Next we consider continuity over (-1,1].

    Let -1<c<1 ,and consider \lim_{x\rightarrow c}{f(x)} =\lim_{x\rightarrow c}{0}= 0 = f(c).

    HENCE f is continuous over (-1,1).

    But if c=1 ,then f(1)=1/2 ,so we see there is a gap from 0 ,which is the value of f(x) for -1<x<1 , to 1/2 =f(1) and hence the function is not continuous.

    To prove that:
    Let, ε=1/4, then for all δ>0 choose ,x=0 ,then |f(x)-f(1)|=|f(0)-f(1)|=|0-1/2|= \frac{1}{2}\geq\frac{1}{4}=\epsilon.

    Thus \lim_{x\rightarrow 1^-}{f(x)}\neq f(1).

    Now to prove : if \lim_{x\rightarrow c}{f(x)} = f(c) ,then \lim_{x\rightarrow c}{\sqrt{f(x)}} =\sqrt{c} ,where f(x)\geq 0,when x belongs to D.

    For that you must consider the : |\sqrt{f(x)}-\sqrt{c}|.................................................. ..................................2

    Now multiply (2) by \frac{|\sqrt{f(x)}+\sqrt{c}|}{|\sqrt{f(x)}+\sqrt{c  }|} and we get:


    \frac{|f(x)-f(c)|}{\sqrt{f(x)}+\sqrt{c}}.

    But  \sqrt{f(x)}+\sqrt{c}\geq\sqrt{c} ,thus :


    \frac{|f(x)-f(c)|}{\sqrt{f(x)}+\sqrt{c}}\leq\frac{|f(x)-f(c)|}{\sqrt{c}}.................................................. .......................................3

    And if we want (3) smaller than ε>0 we must have :

    |f(x)-f(c)|<\sqrt{c}.\epsilon which is possible since f is continuous at c.

    Hence sqrt(f) is continuous at c
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    Quote Originally Posted by noles2188 View Post
    1) Consider the formula f(x)=lim(n-->infinity)((x^n)/(1+x^n)).
    Let D={x:f(x) is an element of R}. Calculate f(x) for all x elements of D and determine where f: D-->R is continuous.
    The answer above is incomplete.
    f(x) = \left\{ {\begin{array}{lr}<br />
   {0,} & {\left| x \right| < 1}  \\   {\frac{1}{2},} & {x = 1}  \\   {1,} & {\left| x \right| > 1}  \\ \end{array} } \right.
    That means that \mathcal{D} = \mathbb{R}\backslash \left\{ { - 1} \right\}.
    It also means that f is continuous on \mathbb{R}\backslash \left\{ {1, - 1} \right\}.

    The mistake above is: \left| x \right| > 1\;,\;\frac{{x^n }}{{1 + x^n }} = \frac{1}<br />
{{\left( {\frac{1}{x}} \right)^n  + 1}}.
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    Quote Originally Posted by Plato View Post
    The mistake above is: \left| x \right| > 1\;,\;\frac{{x^n }}{{1 + x^n }} = \frac{1}<br />
{{\left( {\frac{1}{x}} \right)^n + 1}}.
    If |x|>1 ,then x<-1 or x>1 ,but for x=-2 for example the limit of ;

    \frac{1}{\frac{1}{(-2)^n}+1} is not equal to 1
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    Quote Originally Posted by xalk View Post
    the limit of \frac{1}{\frac{1}{(-2)^n}+1} is not equal to 1
    Yes it is! The reason is that |a|<1\:\Rightarrow\:a^n\to0; and \bigl|\tfrac1{-2}\bigr|<1. So \frac{1}{\frac{1}{(-2)^n}+1}\to\frac1{0+1}=1 as n\to\infty.
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