Showing ${f(frac{1}{n+1})}$ converges in $mathbb{R}$












2












$begingroup$


Question: Let $f:(0,1) to mathbb{R}$ be a differentiable function such that $|f'(x)| leq 5$, for all $x in (0,1)$. Show that the sequence ${f(frac{1}{n+1})}$ converges in $mathbb{R}$.



Attempt:



Consider any $a, b$ such that $[a,b] subset (0,1)$. By LMVT $|frac{f(b)-f(a)}{b-a}| leq 5$ $implies |f(b)-f(a)|leq 5|b-a|$.



The sequence ${frac{1}{n+1}$} converges. Hence, by Cauchy's General Principle, $forall epsilon >0$, $exists N>0$ such that $|frac{1}{n+1}-frac{1}{m+1}| < epsilon/5$ for every $m, n >N$.



Hence, $ |f(frac{1}{n+1})-f(frac{1}{m+1})|leq 5|frac{1}{n+1}-frac{1}{m+1}|< epsilon$, whenever $m, n >N$.



PS: Please do check the solution thoroughly. Kindly correct any "false assumptions", "weak -wordings", "misplaced predicates" etc. A little help goes a long way. I am struggling with proper proof writing right now.






real-analysis calculus proof-verification cauchy-sequences lipschitz-functions







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$endgroup$












  • $begingroup$
    Your answer is Correct.
    $endgroup$
    – hamam_Abdallah
    Jan 12 at 21:02










  • $begingroup$
    Looks fine to.me.
    $endgroup$
    – Peter Szilas
    Jan 12 at 22:09
















2












$begingroup$


Question: Let $f:(0,1) to mathbb{R}$ be a differentiable function such that $|f'(x)| leq 5$, for all $x in (0,1)$. Show that the sequence ${f(frac{1}{n+1})}$ converges in $mathbb{R}$.



Attempt:



Consider any $a, b$ such that $[a,b] subset (0,1)$. By LMVT $|frac{f(b)-f(a)}{b-a}| leq 5$ $implies |f(b)-f(a)|leq 5|b-a|$.



The sequence ${frac{1}{n+1}$} converges. Hence, by Cauchy's General Principle, $forall epsilon >0$, $exists N>0$ such that $|frac{1}{n+1}-frac{1}{m+1}| < epsilon/5$ for every $m, n >N$.



Hence, $ |f(frac{1}{n+1})-f(frac{1}{m+1})|leq 5|frac{1}{n+1}-frac{1}{m+1}|< epsilon$, whenever $m, n >N$.



PS: Please do check the solution thoroughly. Kindly correct any "false assumptions", "weak -wordings", "misplaced predicates" etc. A little help goes a long way. I am struggling with proper proof writing right now.






real-analysis calculus proof-verification cauchy-sequences lipschitz-functions







share|cite|improve this question











$endgroup$












  • $begingroup$
    Your answer is Correct.
    $endgroup$
    – hamam_Abdallah
    Jan 12 at 21:02










  • $begingroup$
    Looks fine to.me.
    $endgroup$
    – Peter Szilas
    Jan 12 at 22:09














2












2








2





$begingroup$


Question: Let $f:(0,1) to mathbb{R}$ be a differentiable function such that $|f'(x)| leq 5$, for all $x in (0,1)$. Show that the sequence ${f(frac{1}{n+1})}$ converges in $mathbb{R}$.



Attempt:



Consider any $a, b$ such that $[a,b] subset (0,1)$. By LMVT $|frac{f(b)-f(a)}{b-a}| leq 5$ $implies |f(b)-f(a)|leq 5|b-a|$.



The sequence ${frac{1}{n+1}$} converges. Hence, by Cauchy's General Principle, $forall epsilon >0$, $exists N>0$ such that $|frac{1}{n+1}-frac{1}{m+1}| < epsilon/5$ for every $m, n >N$.



Hence, $ |f(frac{1}{n+1})-f(frac{1}{m+1})|leq 5|frac{1}{n+1}-frac{1}{m+1}|< epsilon$, whenever $m, n >N$.



PS: Please do check the solution thoroughly. Kindly correct any "false assumptions", "weak -wordings", "misplaced predicates" etc. A little help goes a long way. I am struggling with proper proof writing right now.






real-analysis calculus proof-verification cauchy-sequences lipschitz-functions







share|cite|improve this question











$endgroup$




Question: Let $f:(0,1) to mathbb{R}$ be a differentiable function such that $|f'(x)| leq 5$, for all $x in (0,1)$. Show that the sequence ${f(frac{1}{n+1})}$ converges in $mathbb{R}$.



Attempt:



Consider any $a, b$ such that $[a,b] subset (0,1)$. By LMVT $|frac{f(b)-f(a)}{b-a}| leq 5$ $implies |f(b)-f(a)|leq 5|b-a|$.



The sequence ${frac{1}{n+1}$} converges. Hence, by Cauchy's General Principle, $forall epsilon >0$, $exists N>0$ such that $|frac{1}{n+1}-frac{1}{m+1}| < epsilon/5$ for every $m, n >N$.



Hence, $ |f(frac{1}{n+1})-f(frac{1}{m+1})|leq 5|frac{1}{n+1}-frac{1}{m+1}|< epsilon$, whenever $m, n >N$.



PS: Please do check the solution thoroughly. Kindly correct any "false assumptions", "weak -wordings", "misplaced predicates" etc. A little help goes a long way. I am struggling with proper proof writing right now.






real-analysis calculus proof-verification cauchy-sequences lipschitz-functions




real-analysis calculus proof-verification cauchy-sequences lipschitz-functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 12 at 20:54







Subhasis Biswas

















asked Jan 12 at 20:45









Subhasis BiswasSubhasis Biswas

512411




512411












  • $begingroup$
    Your answer is Correct.
    $endgroup$
    – hamam_Abdallah
    Jan 12 at 21:02










  • $begingroup$
    Looks fine to.me.
    $endgroup$
    – Peter Szilas
    Jan 12 at 22:09


















  • $begingroup$
    Your answer is Correct.
    $endgroup$
    – hamam_Abdallah
    Jan 12 at 21:02










  • $begingroup$
    Looks fine to.me.
    $endgroup$
    – Peter Szilas
    Jan 12 at 22:09
















$begingroup$
Your answer is Correct.
$endgroup$
– hamam_Abdallah
Jan 12 at 21:02




$begingroup$
Your answer is Correct.
$endgroup$
– hamam_Abdallah
Jan 12 at 21:02












$begingroup$
Looks fine to.me.
$endgroup$
– Peter Szilas
Jan 12 at 22:09




$begingroup$
Looks fine to.me.
$endgroup$
– Peter Szilas
Jan 12 at 22:09










1 Answer
1






active

oldest

votes


















2












$begingroup$

An other approach.



$$(forall xin (0,1)) ; |f'(x)|le 5 implies $$



$$f text{ is Uniformly continuous at } (0,1) implies $$



$$lim_{xto 0^+}f(x) text{ exists in } Bbb R implies$$



$$lim_{nto +infty}f(frac{1}{n+1})in Bbb R.$$



PS



I know that tomorrow my total points will decrease with NO reason. should i change my name.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    hamam.Please elaborate.f uniformly continuous in (0,1) implies $ lim_{x rightarrow 0^+} f(x) in R exists.Thanks.
    $endgroup$
    – Peter Szilas
    Jan 12 at 22:22










  • $begingroup$
    C'est le théorème du prolongement des fonctions uniformément continues.
    $endgroup$
    – hamam_Abdallah
    Jan 12 at 22:53










  • $begingroup$
    hamam.Merci bien., Peter
    $endgroup$
    – Peter Szilas
    Jan 13 at 9:20











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









2












$begingroup$

An other approach.



$$(forall xin (0,1)) ; |f'(x)|le 5 implies $$



$$f text{ is Uniformly continuous at } (0,1) implies $$



$$lim_{xto 0^+}f(x) text{ exists in } Bbb R implies$$



$$lim_{nto +infty}f(frac{1}{n+1})in Bbb R.$$



PS



I know that tomorrow my total points will decrease with NO reason. should i change my name.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    hamam.Please elaborate.f uniformly continuous in (0,1) implies $ lim_{x rightarrow 0^+} f(x) in R exists.Thanks.
    $endgroup$
    – Peter Szilas
    Jan 12 at 22:22










  • $begingroup$
    C'est le théorème du prolongement des fonctions uniformément continues.
    $endgroup$
    – hamam_Abdallah
    Jan 12 at 22:53










  • $begingroup$
    hamam.Merci bien., Peter
    $endgroup$
    – Peter Szilas
    Jan 13 at 9:20
















2












$begingroup$

An other approach.



$$(forall xin (0,1)) ; |f'(x)|le 5 implies $$



$$f text{ is Uniformly continuous at } (0,1) implies $$



$$lim_{xto 0^+}f(x) text{ exists in } Bbb R implies$$



$$lim_{nto +infty}f(frac{1}{n+1})in Bbb R.$$



PS



I know that tomorrow my total points will decrease with NO reason. should i change my name.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    hamam.Please elaborate.f uniformly continuous in (0,1) implies $ lim_{x rightarrow 0^+} f(x) in R exists.Thanks.
    $endgroup$
    – Peter Szilas
    Jan 12 at 22:22










  • $begingroup$
    C'est le théorème du prolongement des fonctions uniformément continues.
    $endgroup$
    – hamam_Abdallah
    Jan 12 at 22:53










  • $begingroup$
    hamam.Merci bien., Peter
    $endgroup$
    – Peter Szilas
    Jan 13 at 9:20














2












2








2





$begingroup$

An other approach.



$$(forall xin (0,1)) ; |f'(x)|le 5 implies $$



$$f text{ is Uniformly continuous at } (0,1) implies $$



$$lim_{xto 0^+}f(x) text{ exists in } Bbb R implies$$



$$lim_{nto +infty}f(frac{1}{n+1})in Bbb R.$$



PS



I know that tomorrow my total points will decrease with NO reason. should i change my name.






share|cite|improve this answer









$endgroup$



An other approach.



$$(forall xin (0,1)) ; |f'(x)|le 5 implies $$



$$f text{ is Uniformly continuous at } (0,1) implies $$



$$lim_{xto 0^+}f(x) text{ exists in } Bbb R implies$$



$$lim_{nto +infty}f(frac{1}{n+1})in Bbb R.$$



PS



I know that tomorrow my total points will decrease with NO reason. should i change my name.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 12 at 21:01









hamam_Abdallahhamam_Abdallah

38.2k21634




38.2k21634












  • $begingroup$
    hamam.Please elaborate.f uniformly continuous in (0,1) implies $ lim_{x rightarrow 0^+} f(x) in R exists.Thanks.
    $endgroup$
    – Peter Szilas
    Jan 12 at 22:22










  • $begingroup$
    C'est le théorème du prolongement des fonctions uniformément continues.
    $endgroup$
    – hamam_Abdallah
    Jan 12 at 22:53










  • $begingroup$
    hamam.Merci bien., Peter
    $endgroup$
    – Peter Szilas
    Jan 13 at 9:20


















  • $begingroup$
    hamam.Please elaborate.f uniformly continuous in (0,1) implies $ lim_{x rightarrow 0^+} f(x) in R exists.Thanks.
    $endgroup$
    – Peter Szilas
    Jan 12 at 22:22










  • $begingroup$
    C'est le théorème du prolongement des fonctions uniformément continues.
    $endgroup$
    – hamam_Abdallah
    Jan 12 at 22:53










  • $begingroup$
    hamam.Merci bien., Peter
    $endgroup$
    – Peter Szilas
    Jan 13 at 9:20
















$begingroup$
hamam.Please elaborate.f uniformly continuous in (0,1) implies $ lim_{x rightarrow 0^+} f(x) in R exists.Thanks.
$endgroup$
– Peter Szilas
Jan 12 at 22:22




$begingroup$
hamam.Please elaborate.f uniformly continuous in (0,1) implies $ lim_{x rightarrow 0^+} f(x) in R exists.Thanks.
$endgroup$
– Peter Szilas
Jan 12 at 22:22












$begingroup$
C'est le théorème du prolongement des fonctions uniformément continues.
$endgroup$
– hamam_Abdallah
Jan 12 at 22:53




$begingroup$
C'est le théorème du prolongement des fonctions uniformément continues.
$endgroup$
– hamam_Abdallah
Jan 12 at 22:53












$begingroup$
hamam.Merci bien., Peter
$endgroup$
– Peter Szilas
Jan 13 at 9:20




$begingroup$
hamam.Merci bien., Peter
$endgroup$
– Peter Szilas
Jan 13 at 9:20


















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