# Continuous functions

• April 8th 2009, 09:51 PM
noles2188
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):D-->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.
• April 9th 2009, 07:59 AM
xalk
Quote:

Originally Posted by noles2188
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):D-->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
• April 9th 2009, 03:54 PM
Plato
Quote:

Originally Posted by noles2188
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.

$f(x) = \left\{ {\begin{array}{lr}
{0,} & {\left| x \right| < 1} \\ {\frac{1}{2},} & {x = 1} \\ {1,} & {\left| x \right| > 1} \\ \end{array} } \right.$

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: $\left| x \right| > 1\;,\;\frac{{x^n }}{{1 + x^n }} = \frac{1}
{{\left( {\frac{1}{x}} \right)^n + 1}}$
.
• April 9th 2009, 06:45 PM
xalk
Quote:

Originally Posted by Plato
The mistake above is: $\left| x \right| > 1\;,\;\frac{{x^n }}{{1 + x^n }} = \frac{1}
{{\left( {\frac{1}{x}} \right)^n + 1}}$
.

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
• April 10th 2009, 12:10 AM
Opalg
Quote:

Originally Posted by xalk
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 $n\to\infty$.