Sequence

dwsmith

MHF Hall of Honor
Mar 2010
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\(\displaystyle \forall n\in\mathbb{Z}^+\), let \(\displaystyle f_n\) be the function on the interval from [0,1] by \(\displaystyle f_n(x)=\frac{x^n}{1+x^n}\). Which of the following statements are true?

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

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

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

3. \(\displaystyle \lim_{n\to\infty}\int_0^1f_n(x)dx=\int_0^1(\lim_{n\to\infty}f_n(x))dx\)
Don't know
 
Last edited:
Nov 2009
485
184
\(\displaystyle \forall n\in\mathbb{Z}^+\), let \(\displaystyle f_n\) be the function on the interval from [0,1] by \(\displaystyle f_n(x)=\frac{x^n}{1+x^n}\). Which of the following statements are true?

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

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

dwsmith

MHF Hall of Honor
Mar 2010
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582
Florida
This isn't quite correct. You need to consider two cases: either \(\displaystyle x\geq 1\) or \(\displaystyle x<1\). Try recalculating your limit.

Also, did you have any ideas on the other two?
If x is less than 1, there would be \(\displaystyle \frac{-x}{-x}\) which is still 1. Same situation if x is positive.
 
Nov 2009
485
184
Sorry - I meant to write \(\displaystyle |x|<1\) and \(\displaystyle |x|\geq 1\).

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

MHF Hall of Honor
Mar 2010
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582
Florida
Sorry - I meant to write \(\displaystyle |x|<1\) and \(\displaystyle |x|\geq 1\).

The important bit is this: if \(\displaystyle x\) is fractional, then \(\displaystyle x^n\to 0\). Therefore, \(\displaystyle \frac{x^n}{1+x^n}\to 0\).
If it converges to 0 for all \(\displaystyle 0<x<1\), 1 is still true though.
 
Nov 2009
485
184
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 \(\displaystyle x=-1\) when \(\displaystyle n\) is odd.
 

dwsmith

MHF Hall of Honor
Mar 2010
3,093
582
Florida
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 \(\displaystyle x=-1\) when \(\displaystyle n\) is odd.
What would be an appropriate Mn for two to use?
 
Nov 2009
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184
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 \(\displaystyle \varepsilon>0\), there is \(\displaystyle N\) so that whenever \(\displaystyle n>N\), \(\displaystyle |f_n(x)-f(x)|<\varepsilon\) for all \(\displaystyle x\in [0,1]\). (Now I realise that you are restricted to \(\displaystyle [0,1]\) - sorry for the confusion earlier.) Can you find such an \(\displaystyle N\)? You should take the point \(\displaystyle x=1\) into consideration.