Uniform convergence of power sequence












1












$begingroup$


I'd like to prove that the power sequence



$f_n(x) = x^n$



doesn't converges uniformly on $[0,1]$, but it does on $[0,a]$ if $a < 1$.



My textbook states that a sequence of functions converges uniformly to $f(x)$ if



$ forall , epsilon > 0 ,, exists , overline{n}_{epsilon}: forall x in A quad |f_n(x) - f(x)| < epsilon$



So I found that $f_n(x)$ converges pointwise to
$f(a,b) =
begin{cases}
text{$0 leq x < 1 implies 0$}\
text{$x = 1 ,,,, ,,,,,,implies 1$}
end{cases}
$



So I tried to apply the definition of uniform convergence, but I can't understand how could I find an $epsilon > 0: forall , overline{n}, ,,exists , x in [0,1], ,,, exists , n geq overline{n}: |f_n(x) - f(x)| geq epsilon$



Did I misunderstand something about the definition? Thank you in advance






real-analysis calculus sequences-and-series functions uniform-convergence







share|cite|improve this question











$endgroup$












  • $begingroup$
    How can I construct that sequence?
    $endgroup$
    – Francesco Andreuzzi
    Dec 31 '18 at 12:11










  • $begingroup$
    $f_n(1-1/2) = (1/2)^n$, $,, f(1-1/2) = 0$. But $(1/2)^2 < 1/2$
    $endgroup$
    – Francesco Andreuzzi
    Dec 31 '18 at 12:18












  • $begingroup$
    I take $x_n=1-1/n$, not $1-1/2$. You have $f_n(1-1/n)=(1-1/n)^nto 1/e$ and $f_n(1-1/n)geq frac{1}{4}$ for all $n$ (so take $varepsilon =1/4$ instead of $1/2$).
    $endgroup$
    – Surb
    Dec 31 '18 at 12:28












  • $begingroup$
    Okay, so $x_n to 1/e$ does the trick. Is it ok to take a value of $x$ for $f_n(x)$ that depends on $n$?
    $endgroup$
    – Francesco Andreuzzi
    Dec 31 '18 at 12:38










  • $begingroup$
    I don't understand your question : Is it ok to take a value of x for fn(x) that depends on n? Also, it's not correct that $x_nto 1/e$.
    $endgroup$
    – Surb
    Dec 31 '18 at 12:39


















1












$begingroup$


I'd like to prove that the power sequence



$f_n(x) = x^n$



doesn't converges uniformly on $[0,1]$, but it does on $[0,a]$ if $a < 1$.



My textbook states that a sequence of functions converges uniformly to $f(x)$ if



$ forall , epsilon > 0 ,, exists , overline{n}_{epsilon}: forall x in A quad |f_n(x) - f(x)| < epsilon$



So I found that $f_n(x)$ converges pointwise to
$f(a,b) =
begin{cases}
text{$0 leq x < 1 implies 0$}\
text{$x = 1 ,,,, ,,,,,,implies 1$}
end{cases}
$



So I tried to apply the definition of uniform convergence, but I can't understand how could I find an $epsilon > 0: forall , overline{n}, ,,exists , x in [0,1], ,,, exists , n geq overline{n}: |f_n(x) - f(x)| geq epsilon$



Did I misunderstand something about the definition? Thank you in advance






real-analysis calculus sequences-and-series functions uniform-convergence







share|cite|improve this question











$endgroup$












  • $begingroup$
    How can I construct that sequence?
    $endgroup$
    – Francesco Andreuzzi
    Dec 31 '18 at 12:11










  • $begingroup$
    $f_n(1-1/2) = (1/2)^n$, $,, f(1-1/2) = 0$. But $(1/2)^2 < 1/2$
    $endgroup$
    – Francesco Andreuzzi
    Dec 31 '18 at 12:18












  • $begingroup$
    I take $x_n=1-1/n$, not $1-1/2$. You have $f_n(1-1/n)=(1-1/n)^nto 1/e$ and $f_n(1-1/n)geq frac{1}{4}$ for all $n$ (so take $varepsilon =1/4$ instead of $1/2$).
    $endgroup$
    – Surb
    Dec 31 '18 at 12:28












  • $begingroup$
    Okay, so $x_n to 1/e$ does the trick. Is it ok to take a value of $x$ for $f_n(x)$ that depends on $n$?
    $endgroup$
    – Francesco Andreuzzi
    Dec 31 '18 at 12:38










  • $begingroup$
    I don't understand your question : Is it ok to take a value of x for fn(x) that depends on n? Also, it's not correct that $x_nto 1/e$.
    $endgroup$
    – Surb
    Dec 31 '18 at 12:39
















1












1








1





$begingroup$


I'd like to prove that the power sequence



$f_n(x) = x^n$



doesn't converges uniformly on $[0,1]$, but it does on $[0,a]$ if $a < 1$.



My textbook states that a sequence of functions converges uniformly to $f(x)$ if



$ forall , epsilon > 0 ,, exists , overline{n}_{epsilon}: forall x in A quad |f_n(x) - f(x)| < epsilon$



So I found that $f_n(x)$ converges pointwise to
$f(a,b) =
begin{cases}
text{$0 leq x < 1 implies 0$}\
text{$x = 1 ,,,, ,,,,,,implies 1$}
end{cases}
$



So I tried to apply the definition of uniform convergence, but I can't understand how could I find an $epsilon > 0: forall , overline{n}, ,,exists , x in [0,1], ,,, exists , n geq overline{n}: |f_n(x) - f(x)| geq epsilon$



Did I misunderstand something about the definition? Thank you in advance






real-analysis calculus sequences-and-series functions uniform-convergence







share|cite|improve this question











$endgroup$




I'd like to prove that the power sequence



$f_n(x) = x^n$



doesn't converges uniformly on $[0,1]$, but it does on $[0,a]$ if $a < 1$.



My textbook states that a sequence of functions converges uniformly to $f(x)$ if



$ forall , epsilon > 0 ,, exists , overline{n}_{epsilon}: forall x in A quad |f_n(x) - f(x)| < epsilon$



So I found that $f_n(x)$ converges pointwise to
$f(a,b) =
begin{cases}
text{$0 leq x < 1 implies 0$}\
text{$x = 1 ,,,, ,,,,,,implies 1$}
end{cases}
$



So I tried to apply the definition of uniform convergence, but I can't understand how could I find an $epsilon > 0: forall , overline{n}, ,,exists , x in [0,1], ,,, exists , n geq overline{n}: |f_n(x) - f(x)| geq epsilon$



Did I misunderstand something about the definition? Thank you in advance






real-analysis calculus sequences-and-series functions uniform-convergence




real-analysis calculus sequences-and-series functions uniform-convergence






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 31 '18 at 13:44







Francesco Andreuzzi

















asked Dec 31 '18 at 11:55









Francesco AndreuzziFrancesco Andreuzzi

63




63












  • $begingroup$
    How can I construct that sequence?
    $endgroup$
    – Francesco Andreuzzi
    Dec 31 '18 at 12:11










  • $begingroup$
    $f_n(1-1/2) = (1/2)^n$, $,, f(1-1/2) = 0$. But $(1/2)^2 < 1/2$
    $endgroup$
    – Francesco Andreuzzi
    Dec 31 '18 at 12:18












  • $begingroup$
    I take $x_n=1-1/n$, not $1-1/2$. You have $f_n(1-1/n)=(1-1/n)^nto 1/e$ and $f_n(1-1/n)geq frac{1}{4}$ for all $n$ (so take $varepsilon =1/4$ instead of $1/2$).
    $endgroup$
    – Surb
    Dec 31 '18 at 12:28












  • $begingroup$
    Okay, so $x_n to 1/e$ does the trick. Is it ok to take a value of $x$ for $f_n(x)$ that depends on $n$?
    $endgroup$
    – Francesco Andreuzzi
    Dec 31 '18 at 12:38










  • $begingroup$
    I don't understand your question : Is it ok to take a value of x for fn(x) that depends on n? Also, it's not correct that $x_nto 1/e$.
    $endgroup$
    – Surb
    Dec 31 '18 at 12:39




















  • $begingroup$
    How can I construct that sequence?
    $endgroup$
    – Francesco Andreuzzi
    Dec 31 '18 at 12:11










  • $begingroup$
    $f_n(1-1/2) = (1/2)^n$, $,, f(1-1/2) = 0$. But $(1/2)^2 < 1/2$
    $endgroup$
    – Francesco Andreuzzi
    Dec 31 '18 at 12:18












  • $begingroup$
    I take $x_n=1-1/n$, not $1-1/2$. You have $f_n(1-1/n)=(1-1/n)^nto 1/e$ and $f_n(1-1/n)geq frac{1}{4}$ for all $n$ (so take $varepsilon =1/4$ instead of $1/2$).
    $endgroup$
    – Surb
    Dec 31 '18 at 12:28












  • $begingroup$
    Okay, so $x_n to 1/e$ does the trick. Is it ok to take a value of $x$ for $f_n(x)$ that depends on $n$?
    $endgroup$
    – Francesco Andreuzzi
    Dec 31 '18 at 12:38










  • $begingroup$
    I don't understand your question : Is it ok to take a value of x for fn(x) that depends on n? Also, it's not correct that $x_nto 1/e$.
    $endgroup$
    – Surb
    Dec 31 '18 at 12:39


















$begingroup$
How can I construct that sequence?
$endgroup$
– Francesco Andreuzzi
Dec 31 '18 at 12:11




$begingroup$
How can I construct that sequence?
$endgroup$
– Francesco Andreuzzi
Dec 31 '18 at 12:11












$begingroup$
$f_n(1-1/2) = (1/2)^n$, $,, f(1-1/2) = 0$. But $(1/2)^2 < 1/2$
$endgroup$
– Francesco Andreuzzi
Dec 31 '18 at 12:18






$begingroup$
$f_n(1-1/2) = (1/2)^n$, $,, f(1-1/2) = 0$. But $(1/2)^2 < 1/2$
$endgroup$
– Francesco Andreuzzi
Dec 31 '18 at 12:18














$begingroup$
I take $x_n=1-1/n$, not $1-1/2$. You have $f_n(1-1/n)=(1-1/n)^nto 1/e$ and $f_n(1-1/n)geq frac{1}{4}$ for all $n$ (so take $varepsilon =1/4$ instead of $1/2$).
$endgroup$
– Surb
Dec 31 '18 at 12:28






$begingroup$
I take $x_n=1-1/n$, not $1-1/2$. You have $f_n(1-1/n)=(1-1/n)^nto 1/e$ and $f_n(1-1/n)geq frac{1}{4}$ for all $n$ (so take $varepsilon =1/4$ instead of $1/2$).
$endgroup$
– Surb
Dec 31 '18 at 12:28














$begingroup$
Okay, so $x_n to 1/e$ does the trick. Is it ok to take a value of $x$ for $f_n(x)$ that depends on $n$?
$endgroup$
– Francesco Andreuzzi
Dec 31 '18 at 12:38




$begingroup$
Okay, so $x_n to 1/e$ does the trick. Is it ok to take a value of $x$ for $f_n(x)$ that depends on $n$?
$endgroup$
– Francesco Andreuzzi
Dec 31 '18 at 12:38












$begingroup$
I don't understand your question : Is it ok to take a value of x for fn(x) that depends on n? Also, it's not correct that $x_nto 1/e$.
$endgroup$
– Surb
Dec 31 '18 at 12:39






$begingroup$
I don't understand your question : Is it ok to take a value of x for fn(x) that depends on n? Also, it's not correct that $x_nto 1/e$.
$endgroup$
– Surb
Dec 31 '18 at 12:39












1 Answer
1






active

oldest

votes


















1












$begingroup$

In the way that I think is more constructive to think about it, uniform convergence of $f_n$ to $f$ on $A$ is equivalent to $underset{xin A}{sup} vert f_n(x)-f(x)vert overset{nrightarrow infty}{rightarrow} 0$.
So if for all $n$ there exists $x_nin A$ such that:



$vert f_n(x_n)-f(x_n)vertequiv c>0$



then you would show that there is no uniform to $f$. You have to consider just one $f$, since uniform convergence also implies point-wise convergence.



You could have also gone another route. If a sequence of continuous functions converges uniformly to $f$, then $f$ is a continuous function, which you can see that $f$ is not.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Could you please explain which $x_n$ should I consider?
    $endgroup$
    – Francesco Andreuzzi
    Dec 31 '18 at 12:45










  • $begingroup$
    For example if you want to show for $c=frac{3}{4}$, you would want that $x_n^n-0=frac{3}{4}$. So for example $x_n:=Big( frac{3}{4} Big)^{frac{1}{n}} $ would work.
    $endgroup$
    – Keen-ameteur
    Dec 31 '18 at 12:52












  • $begingroup$
    But I know that $f_n(x) = x^n$ converges on $[0,a]$ if $a < 1$. Your proof proves also that my sequence doesn't converge uniformly on that set
    $endgroup$
    – Francesco Andreuzzi
    Dec 31 '18 at 13:01










  • $begingroup$
    It won't since for any $1>a>0$, $x_n$ would eventually be larger than $a$, and thus not be relevant to the interval $[0,a]$. Furthermore for all $c>0$, choosing $x_n =sqrt[n]{c}$ would mean that $x_n>a$ eventually.
    $endgroup$
    – Keen-ameteur
    Dec 31 '18 at 13:04












  • $begingroup$
    Okay, it makes sense. Thank you very much. Could you please check my last comment below my question, and tell me what you think? For instance, $lim_{n to infty} (3/4)^{1/n} = 1$, but this conflicts with the pointwise limit that I definied in my question
    $endgroup$
    – Francesco Andreuzzi
    Dec 31 '18 at 13:09













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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

In the way that I think is more constructive to think about it, uniform convergence of $f_n$ to $f$ on $A$ is equivalent to $underset{xin A}{sup} vert f_n(x)-f(x)vert overset{nrightarrow infty}{rightarrow} 0$.
So if for all $n$ there exists $x_nin A$ such that:



$vert f_n(x_n)-f(x_n)vertequiv c>0$



then you would show that there is no uniform to $f$. You have to consider just one $f$, since uniform convergence also implies point-wise convergence.



You could have also gone another route. If a sequence of continuous functions converges uniformly to $f$, then $f$ is a continuous function, which you can see that $f$ is not.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Could you please explain which $x_n$ should I consider?
    $endgroup$
    – Francesco Andreuzzi
    Dec 31 '18 at 12:45










  • $begingroup$
    For example if you want to show for $c=frac{3}{4}$, you would want that $x_n^n-0=frac{3}{4}$. So for example $x_n:=Big( frac{3}{4} Big)^{frac{1}{n}} $ would work.
    $endgroup$
    – Keen-ameteur
    Dec 31 '18 at 12:52












  • $begingroup$
    But I know that $f_n(x) = x^n$ converges on $[0,a]$ if $a < 1$. Your proof proves also that my sequence doesn't converge uniformly on that set
    $endgroup$
    – Francesco Andreuzzi
    Dec 31 '18 at 13:01










  • $begingroup$
    It won't since for any $1>a>0$, $x_n$ would eventually be larger than $a$, and thus not be relevant to the interval $[0,a]$. Furthermore for all $c>0$, choosing $x_n =sqrt[n]{c}$ would mean that $x_n>a$ eventually.
    $endgroup$
    – Keen-ameteur
    Dec 31 '18 at 13:04












  • $begingroup$
    Okay, it makes sense. Thank you very much. Could you please check my last comment below my question, and tell me what you think? For instance, $lim_{n to infty} (3/4)^{1/n} = 1$, but this conflicts with the pointwise limit that I definied in my question
    $endgroup$
    – Francesco Andreuzzi
    Dec 31 '18 at 13:09


















1












$begingroup$

In the way that I think is more constructive to think about it, uniform convergence of $f_n$ to $f$ on $A$ is equivalent to $underset{xin A}{sup} vert f_n(x)-f(x)vert overset{nrightarrow infty}{rightarrow} 0$.
So if for all $n$ there exists $x_nin A$ such that:



$vert f_n(x_n)-f(x_n)vertequiv c>0$



then you would show that there is no uniform to $f$. You have to consider just one $f$, since uniform convergence also implies point-wise convergence.



You could have also gone another route. If a sequence of continuous functions converges uniformly to $f$, then $f$ is a continuous function, which you can see that $f$ is not.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Could you please explain which $x_n$ should I consider?
    $endgroup$
    – Francesco Andreuzzi
    Dec 31 '18 at 12:45










  • $begingroup$
    For example if you want to show for $c=frac{3}{4}$, you would want that $x_n^n-0=frac{3}{4}$. So for example $x_n:=Big( frac{3}{4} Big)^{frac{1}{n}} $ would work.
    $endgroup$
    – Keen-ameteur
    Dec 31 '18 at 12:52












  • $begingroup$
    But I know that $f_n(x) = x^n$ converges on $[0,a]$ if $a < 1$. Your proof proves also that my sequence doesn't converge uniformly on that set
    $endgroup$
    – Francesco Andreuzzi
    Dec 31 '18 at 13:01










  • $begingroup$
    It won't since for any $1>a>0$, $x_n$ would eventually be larger than $a$, and thus not be relevant to the interval $[0,a]$. Furthermore for all $c>0$, choosing $x_n =sqrt[n]{c}$ would mean that $x_n>a$ eventually.
    $endgroup$
    – Keen-ameteur
    Dec 31 '18 at 13:04












  • $begingroup$
    Okay, it makes sense. Thank you very much. Could you please check my last comment below my question, and tell me what you think? For instance, $lim_{n to infty} (3/4)^{1/n} = 1$, but this conflicts with the pointwise limit that I definied in my question
    $endgroup$
    – Francesco Andreuzzi
    Dec 31 '18 at 13:09
















1












1








1





$begingroup$

In the way that I think is more constructive to think about it, uniform convergence of $f_n$ to $f$ on $A$ is equivalent to $underset{xin A}{sup} vert f_n(x)-f(x)vert overset{nrightarrow infty}{rightarrow} 0$.
So if for all $n$ there exists $x_nin A$ such that:



$vert f_n(x_n)-f(x_n)vertequiv c>0$



then you would show that there is no uniform to $f$. You have to consider just one $f$, since uniform convergence also implies point-wise convergence.



You could have also gone another route. If a sequence of continuous functions converges uniformly to $f$, then $f$ is a continuous function, which you can see that $f$ is not.






share|cite|improve this answer











$endgroup$



In the way that I think is more constructive to think about it, uniform convergence of $f_n$ to $f$ on $A$ is equivalent to $underset{xin A}{sup} vert f_n(x)-f(x)vert overset{nrightarrow infty}{rightarrow} 0$.
So if for all $n$ there exists $x_nin A$ such that:



$vert f_n(x_n)-f(x_n)vertequiv c>0$



then you would show that there is no uniform to $f$. You have to consider just one $f$, since uniform convergence also implies point-wise convergence.



You could have also gone another route. If a sequence of continuous functions converges uniformly to $f$, then $f$ is a continuous function, which you can see that $f$ is not.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 31 '18 at 12:54

























answered Dec 31 '18 at 12:36









Keen-ameteurKeen-ameteur

1,477316




1,477316












  • $begingroup$
    Could you please explain which $x_n$ should I consider?
    $endgroup$
    – Francesco Andreuzzi
    Dec 31 '18 at 12:45










  • $begingroup$
    For example if you want to show for $c=frac{3}{4}$, you would want that $x_n^n-0=frac{3}{4}$. So for example $x_n:=Big( frac{3}{4} Big)^{frac{1}{n}} $ would work.
    $endgroup$
    – Keen-ameteur
    Dec 31 '18 at 12:52












  • $begingroup$
    But I know that $f_n(x) = x^n$ converges on $[0,a]$ if $a < 1$. Your proof proves also that my sequence doesn't converge uniformly on that set
    $endgroup$
    – Francesco Andreuzzi
    Dec 31 '18 at 13:01










  • $begingroup$
    It won't since for any $1>a>0$, $x_n$ would eventually be larger than $a$, and thus not be relevant to the interval $[0,a]$. Furthermore for all $c>0$, choosing $x_n =sqrt[n]{c}$ would mean that $x_n>a$ eventually.
    $endgroup$
    – Keen-ameteur
    Dec 31 '18 at 13:04












  • $begingroup$
    Okay, it makes sense. Thank you very much. Could you please check my last comment below my question, and tell me what you think? For instance, $lim_{n to infty} (3/4)^{1/n} = 1$, but this conflicts with the pointwise limit that I definied in my question
    $endgroup$
    – Francesco Andreuzzi
    Dec 31 '18 at 13:09




















  • $begingroup$
    Could you please explain which $x_n$ should I consider?
    $endgroup$
    – Francesco Andreuzzi
    Dec 31 '18 at 12:45










  • $begingroup$
    For example if you want to show for $c=frac{3}{4}$, you would want that $x_n^n-0=frac{3}{4}$. So for example $x_n:=Big( frac{3}{4} Big)^{frac{1}{n}} $ would work.
    $endgroup$
    – Keen-ameteur
    Dec 31 '18 at 12:52












  • $begingroup$
    But I know that $f_n(x) = x^n$ converges on $[0,a]$ if $a < 1$. Your proof proves also that my sequence doesn't converge uniformly on that set
    $endgroup$
    – Francesco Andreuzzi
    Dec 31 '18 at 13:01










  • $begingroup$
    It won't since for any $1>a>0$, $x_n$ would eventually be larger than $a$, and thus not be relevant to the interval $[0,a]$. Furthermore for all $c>0$, choosing $x_n =sqrt[n]{c}$ would mean that $x_n>a$ eventually.
    $endgroup$
    – Keen-ameteur
    Dec 31 '18 at 13:04












  • $begingroup$
    Okay, it makes sense. Thank you very much. Could you please check my last comment below my question, and tell me what you think? For instance, $lim_{n to infty} (3/4)^{1/n} = 1$, but this conflicts with the pointwise limit that I definied in my question
    $endgroup$
    – Francesco Andreuzzi
    Dec 31 '18 at 13:09


















$begingroup$
Could you please explain which $x_n$ should I consider?
$endgroup$
– Francesco Andreuzzi
Dec 31 '18 at 12:45




$begingroup$
Could you please explain which $x_n$ should I consider?
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– Francesco Andreuzzi
Dec 31 '18 at 12:45












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For example if you want to show for $c=frac{3}{4}$, you would want that $x_n^n-0=frac{3}{4}$. So for example $x_n:=Big( frac{3}{4} Big)^{frac{1}{n}} $ would work.
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– Keen-ameteur
Dec 31 '18 at 12:52






$begingroup$
For example if you want to show for $c=frac{3}{4}$, you would want that $x_n^n-0=frac{3}{4}$. So for example $x_n:=Big( frac{3}{4} Big)^{frac{1}{n}} $ would work.
$endgroup$
– Keen-ameteur
Dec 31 '18 at 12:52














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But I know that $f_n(x) = x^n$ converges on $[0,a]$ if $a < 1$. Your proof proves also that my sequence doesn't converge uniformly on that set
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– Francesco Andreuzzi
Dec 31 '18 at 13:01




$begingroup$
But I know that $f_n(x) = x^n$ converges on $[0,a]$ if $a < 1$. Your proof proves also that my sequence doesn't converge uniformly on that set
$endgroup$
– Francesco Andreuzzi
Dec 31 '18 at 13:01












$begingroup$
It won't since for any $1>a>0$, $x_n$ would eventually be larger than $a$, and thus not be relevant to the interval $[0,a]$. Furthermore for all $c>0$, choosing $x_n =sqrt[n]{c}$ would mean that $x_n>a$ eventually.
$endgroup$
– Keen-ameteur
Dec 31 '18 at 13:04






$begingroup$
It won't since for any $1>a>0$, $x_n$ would eventually be larger than $a$, and thus not be relevant to the interval $[0,a]$. Furthermore for all $c>0$, choosing $x_n =sqrt[n]{c}$ would mean that $x_n>a$ eventually.
$endgroup$
– Keen-ameteur
Dec 31 '18 at 13:04














$begingroup$
Okay, it makes sense. Thank you very much. Could you please check my last comment below my question, and tell me what you think? For instance, $lim_{n to infty} (3/4)^{1/n} = 1$, but this conflicts with the pointwise limit that I definied in my question
$endgroup$
– Francesco Andreuzzi
Dec 31 '18 at 13:09






$begingroup$
Okay, it makes sense. Thank you very much. Could you please check my last comment below my question, and tell me what you think? For instance, $lim_{n to infty} (3/4)^{1/n} = 1$, but this conflicts with the pointwise limit that I definied in my question
$endgroup$
– Francesco Andreuzzi
Dec 31 '18 at 13:09




















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