Compute pointwise limit of $f_{n}(x) = 2^nx 0leq x leq frac{1}{2^n}, 1 frac{1}{2^n} < x < 1 $Pointwise...
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Compute pointwise limit of $f_{n}(x) = 2^nx 0leq x leq frac{1}{2^n}, 1 frac{1}{2^n}
Pointwise limit of continuous functions not Riemann integrableDid I compute the limit of of the sequence $x_{n+1}=frac{x_n}{x_n+1}, x_o=1$ properly?If $a_n>0$ and $ displaystyle frac{a_{n+1}}{a_n} leq 1 + frac{1}{n^2}$ and $displaystyle {frac{a_{n+1}}{a_n} } notto 1$ then $a_n to 0$Proving the limit of a recursive sequenceseries of uniformly continuous functions which converge pointwise to the limitProving that $(f_k)_k$ converges pointwise to zero on $[0,1]$Pointwise limitConstruct $(y_n)$ such that $x_n leq y_n leq 2y_{f(n)}$ where $f:mathbb{N}_{geq 0} to mathbb{N}_{geq 1}$ is strictly increasingWhat is the correct proof that when $epsilon$ decreases (in a limit) then the largest $delta$ should also decrease?Finding and proving the limit of a recursive sequence
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Since $ 0leq x/leq frac {1}{2^n}$, $2^n$ should be a decreasing function and clearly decreases to one $forall x in [0,frac{1}{2^n}],$ but how should I go about proving that the function limit is 1?
real-analysis
edited Mar 9 at 17:15
Robert Shore
2,567117
asked Mar 9 at 17:06
Brandon EvansBrandon Evans
243
$endgroup$
add a comment |
$begingroup$
Since $ 0leq x/leq frac {1}{2^n}$, $2^n$ should be a decreasing function and clearly decreases to one $forall x in [0,frac{1}{2^n}],$ but how should I go about proving that the function limit is 1?
real-analysis
edited Mar 9 at 17:15
Robert Shore
2,567117
asked Mar 9 at 17:06
Brandon EvansBrandon Evans
243
$endgroup$
add a comment |
$begingroup$
Since $ 0leq x/leq frac {1}{2^n}$, $2^n$ should be a decreasing function and clearly decreases to one $forall x in [0,frac{1}{2^n}],$ but how should I go about proving that the function limit is 1?
real-analysis
edited Mar 9 at 17:15
Robert Shore
2,567117
asked Mar 9 at 17:06
Brandon EvansBrandon Evans
243
$endgroup$
Since $ 0leq x/leq frac {1}{2^n}$, $2^n$ should be a decreasing function and clearly decreases to one $forall x in [0,frac{1}{2^n}],$ but how should I go about proving that the function limit is 1?
real-analysis
real-analysis
edited Mar 9 at 17:15
Robert Shore
2,567117
asked Mar 9 at 17:06
Brandon EvansBrandon Evans
243
edited Mar 9 at 17:15
Robert Shore
2,567117
asked Mar 9 at 17:06
Brandon EvansBrandon Evans
243
edited Mar 9 at 17:15
Robert Shore
2,567117
edited Mar 9 at 17:15
Robert Shore
2,567117
edited Mar 9 at 17:15
Robert Shore
2,567117
2,567117
asked Mar 9 at 17:06
Brandon EvansBrandon Evans
243
asked Mar 9 at 17:06
Brandon EvansBrandon Evans
243
asked Mar 9 at 17:06
Brandon EvansBrandon Evans
243
243
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Given a $xin(0,1]$, taking $n$ sufficiently large we have that$$frac{1}{2^n}<x,$$
therefore $f(x)=lim_{nrightarrow infty}f_n(x)=1$. But if $x=0$, then $x<frac{1}{2^n}$ for all $ninmathbb{N}$; which leads to $f(0)=0$.
answered Mar 9 at 17:23
Jonathan HonórioJonathan Honório
113
New contributor
Jonathan Honório is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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$begingroup$
Given a $xin(0,1]$, taking $n$ sufficiently large we have that$$frac{1}{2^n}<x,$$
therefore $f(x)=lim_{nrightarrow infty}f_n(x)=1$. But if $x=0$, then $x<frac{1}{2^n}$ for all $ninmathbb{N}$; which leads to $f(0)=0$.
answered Mar 9 at 17:23
Jonathan HonórioJonathan Honório
113
New contributor
Jonathan Honório is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Given a $xin(0,1]$, taking $n$ sufficiently large we have that$$frac{1}{2^n}<x,$$
therefore $f(x)=lim_{nrightarrow infty}f_n(x)=1$. But if $x=0$, then $x<frac{1}{2^n}$ for all $ninmathbb{N}$; which leads to $f(0)=0$.
answered Mar 9 at 17:23
Jonathan HonórioJonathan Honório
113
New contributor
Jonathan Honório is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Given a $xin(0,1]$, taking $n$ sufficiently large we have that$$frac{1}{2^n}<x,$$
therefore $f(x)=lim_{nrightarrow infty}f_n(x)=1$. But if $x=0$, then $x<frac{1}{2^n}$ for all $ninmathbb{N}$; which leads to $f(0)=0$.
answered Mar 9 at 17:23
Jonathan HonórioJonathan Honório
113
New contributor
Jonathan Honório is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Given a $xin(0,1]$, taking $n$ sufficiently large we have that$$frac{1}{2^n}<x,$$
therefore $f(x)=lim_{nrightarrow infty}f_n(x)=1$. But if $x=0$, then $x<frac{1}{2^n}$ for all $ninmathbb{N}$; which leads to $f(0)=0$.
answered Mar 9 at 17:23
Jonathan HonórioJonathan Honório
113
New contributor
Jonathan Honório is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Mar 9 at 17:23
Jonathan HonórioJonathan Honório
113
New contributor
Jonathan Honório is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Mar 9 at 17:23
Jonathan HonórioJonathan Honório
113
answered Mar 9 at 17:23
Jonathan HonórioJonathan Honório
113
113
New contributor
Jonathan Honório is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Jonathan Honório is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Jonathan Honório is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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