1. ## Sequence

$\forall n\in\mathbb{Z}^+$, let $f_n$ be the function on the interval from [0,1] by $f_n(x)=\frac{x^n}{1+x^n}$. Which of the following statements are true?

1. The sequence { $f_n$} converges pointwise on [0,1] to a limit function $f$.
$\lim_{n\to\infty}\frac{x^n}{1+x^n}=\lim_{n\to\inft y}\frac{nx^{n-1}}{nx^{n-1}}=0=f(x)$

$\lim_{n\to\infty}\frac{1^n}{1+1^n}=\frac{1}{2}=f(x )$
True

2. The sequence { $f_n$} converges uniformly on [0,1] to a limit function $f$.
False

3. $\lim_{n\to\infty}\int_0^1f_n(x)dx=\int_0^1(\lim_{n \to\infty}f_n(x))dx$
Don't know

2. Originally Posted by dwsmith
$\forall n\in\mathbb{Z}^+$, let $f_n$ be the function on the interval from [0,1] by $f_n(x)=\frac{x^n}{1+x^n}$. Which of the following statements are true?

1. The sequence { $f_n$} converges pointwise on [0,1] to a limit function $f$.
$\lim_{n\to\infty}\frac{x^n}{1+x^n}=\lim_{n\to\inft y}\frac{nx^{n-1}}{nx^{n-1}}=1$
True
This isn't quite correct. You need to consider two cases: either $x\geq 1$ or $x<1$. Try recalculating your limit.

Also, did you have any ideas on the other two?

3. Originally Posted by roninpro
This isn't quite correct. You need to consider two cases: either $x\geq 1$ or $x<1$. Try recalculating your limit.

Also, did you have any ideas on the other two?
If x is less than 1, there would be $\frac{-x}{-x}$ which is still 1. Same situation if x is positive.

4. Sorry - I meant to write $|x|<1$ and $|x|\geq 1$.

The important bit is this: if $x$ is fractional, then $x^n\to 0$. Therefore, $\frac{x^n}{1+x^n}\to 0$.

5. Originally Posted by roninpro
Sorry - I meant to write $|x|<1$ and $|x|\geq 1$.

The important bit is this: if $x$ is fractional, then $x^n\to 0$. Therefore, $\frac{x^n}{1+x^n}\to 0$.
If it converges to 0 for all $0, 1 is still true though.

6. Yes, that's right. But your limit was incorrect and will affect the way you do part two and three.

For part two, you might try considering that there is an asymptote at $x=-1$ when $n$ is odd.

7. Originally Posted by roninpro
Yes, that's right. But your limit was incorrect and will affect the way you do part two and three.

For part two, you might try considering that there is an asymptote at $x=-1$ when $n$ is odd.
What would be an appropriate Mn for two to use?

8. What do you mean by Mn?

9. Originally Posted by roninpro
What do you mean by Mn?
$|f_n(x)|\leq M_n \ \forall x$

10. Are you trying to use the Weierstrass M-Test? It isn't applicable here since we are not looking at a function defined by an infinite series.

You need to use the definition of uniform convergence: Given any $\varepsilon>0$, there is $N$ so that whenever $n>N$, $|f_n(x)-f(x)|<\varepsilon$ for all $x\in [0,1]$. (Now I realise that you are restricted to $[0,1]$ - sorry for the confusion earlier.) Can you find such an $N$? You should take the point $x=1$ into consideration.

11. Originally Posted by roninpro
Are you trying to use the Weierstrass M-Test? It isn't applicable here since we are not looking at a function defined by an infinite series.

You need to use the definition of uniform convergence: Given any $\varepsilon>0$, there is $N$ so that whenever $n>N$, $|f_n(x)-f(x)|<\varepsilon$ for all $x\in [0,1]$. (Now I realise that you are restricted to $[0,1]$ - sorry for the confusion earlier.) Can you find such an $N$? You should take the point $x=1$ into consideration.
If $x=1$, then $f_n(1)=\frac{1}{2}$.

$|\frac{1}{2}-0|<\varepsilon$
False.

12. This is the right idea, but still not exactly the case. The function converges pointwise to $

f(x) = \begin{cases}

0 & \text{for x<1} \\

\frac{1}{2} & \text{for x=1}
\end{cases}$
. So, it turns out for all $n$, $|f_n(1)-f(1)|=0<\varepsilon$.

Have a look at a plot of the sequence below and give it another shot.

13. Originally Posted by roninpro
This is the right idea, but still not exactly the case. The function converges pointwise to $

f(x) = \begin{cases}

0 & \text{for x<1} \\

1 & \text{for x=1}
\end{cases}$
. So, it turns out for all $n$, $|f_n(1)-f(1)|=0<\varepsilon$.

Have a look at a plot of the sequence below and give it another shot.
If x<1, then the limit is 0 and if x=1, then the limit is .5.

x=1
$|\frac{1}{2}-\frac{1}{2}|=0<\varepsilon$

[0,1)
$|[0,1)-0|=[0,1)<\varepsilon$
which isn't always true.

Hence, 2 is false?

14. Yes, part two is false.

Your argument needs to be refined a little bit: you need to find a point that "floats" away from our function for any $n$. Since all of the $f_n$ are continuous, there exists a point $y_n\in [0,1]$ so that $f_n(y_n)=\frac{1}{4}$. Since this is the case, we have $|f_n(y_n)-f(y_n)|=|\frac{1}{4}-0|=\frac{1}{4}$. Therefore, we can never find $n$ large enough so that $|f_n(x)-f(x)|<\varepsilon=\frac{1}{4}$ for all $x\in [0,1]$. (i.e. Uniform continuity fails if we try to choose $\varepsilon=\frac{1}{4}$.)

In part three, try actually evaluating the integral $\lim_{n\to \infty} \int_0^1 \frac{x^n}{1+x^n} \text{ d}x$ by putting a convenient bound on it.