How this limit equals zero? [closed]

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Example $7.6$ in Rudin's Analysis:

Let $f_n(x)=n^2x(1-x^2)^n$ on $[0,1]$.

For $0<xleq 1$, we have $$lim_n rightarrow infty f_n(x)=0$$ by theorem $3.20 (d)$.

My question is:

How this limit follows from theorem $3.20 (d)$ ?

$textbfReference (Theorem 3.20 (d) ):$

If $p>0$ and $alpha $ is real, then $$lim_n rightarrow infty fracn^alpha(1+p)^n=0$$

real-analysis

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asked Jul 29 at 8:52

Learning Mathematics

464213

closed as off-topic by Did, Xander Henderson, Shailesh, John Ma, amWhy Jul 30 at 0:32

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Xander Henderson, Shailesh, John Ma, amWhy

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  Make $1-x^2=1/(1+p)$.
  – Yves Daoust
  Jul 29 at 8:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Have you properly understood why this theorem holds good?
  – Arnab Chowdhury
  Jul 29 at 9:12
  

  
  
  
  

add a comment | 

up vote
-2
down vote

favorite

Example $7.6$ in Rudin's Analysis:

Let $f_n(x)=n^2x(1-x^2)^n$ on $[0,1]$.

For $0<xleq 1$, we have $$lim_n rightarrow infty f_n(x)=0$$ by theorem $3.20 (d)$.

My question is:

How this limit follows from theorem $3.20 (d)$ ?

$textbfReference (Theorem 3.20 (d) ):$

If $p>0$ and $alpha $ is real, then $$lim_n rightarrow infty fracn^alpha(1+p)^n=0$$

real-analysis

share|cite|improve this question

asked Jul 29 at 8:52

Learning Mathematics

464213

closed as off-topic by Did, Xander Henderson, Shailesh, John Ma, amWhy Jul 30 at 0:32

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Xander Henderson, Shailesh, John Ma, amWhy

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  Make $1-x^2=1/(1+p)$.
  – Yves Daoust
  Jul 29 at 8:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Have you properly understood why this theorem holds good?
  – Arnab Chowdhury
  Jul 29 at 9:12
  

  
  
  
  

add a comment | 

up vote
-2
down vote

favorite

up vote
-2
down vote

favorite

Example $7.6$ in Rudin's Analysis:

Let $f_n(x)=n^2x(1-x^2)^n$ on $[0,1]$.

For $0<xleq 1$, we have $$lim_n rightarrow infty f_n(x)=0$$ by theorem $3.20 (d)$.

My question is:

How this limit follows from theorem $3.20 (d)$ ?

$textbfReference (Theorem 3.20 (d) ):$

If $p>0$ and $alpha $ is real, then $$lim_n rightarrow infty fracn^alpha(1+p)^n=0$$

real-analysis

share|cite|improve this question

asked Jul 29 at 8:52

Learning Mathematics

464213

Example $7.6$ in Rudin's Analysis:

Let $f_n(x)=n^2x(1-x^2)^n$ on $[0,1]$.

For $0<xleq 1$, we have $$lim_n rightarrow infty f_n(x)=0$$ by theorem $3.20 (d)$.

My question is:

How this limit follows from theorem $3.20 (d)$ ?

$textbfReference (Theorem 3.20 (d) ):$

If $p>0$ and $alpha $ is real, then $$lim_n rightarrow infty fracn^alpha(1+p)^n=0$$

real-analysis

share|cite|improve this question

asked Jul 29 at 8:52

Learning Mathematics

464213

share|cite|improve this question

share|cite|improve this question

asked Jul 29 at 8:52

Learning Mathematics

464213

asked Jul 29 at 8:52

Learning Mathematics

464213

asked Jul 29 at 8:52

Learning Mathematics

464213

464213

closed as off-topic by Did, Xander Henderson, Shailesh, John Ma, amWhy Jul 30 at 0:32

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Xander Henderson, Shailesh, John Ma, amWhy

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by Did, Xander Henderson, Shailesh, John Ma, amWhy Jul 30 at 0:32

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Xander Henderson, Shailesh, John Ma, amWhy

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  Make $1-x^2=1/(1+p)$.
  – Yves Daoust
  Jul 29 at 8:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Have you properly understood why this theorem holds good?
  – Arnab Chowdhury
  Jul 29 at 9:12
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  Make $1-x^2=1/(1+p)$.
  – Yves Daoust
  Jul 29 at 8:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Have you properly understood why this theorem holds good?
  – Arnab Chowdhury
  Jul 29 at 9:12
  

  
  
  
  

4

4

Make $1-x^2=1/(1+p)$.
– Yves Daoust
Jul 29 at 8:55

Make $1-x^2=1/(1+p)$.
– Yves Daoust
Jul 29 at 8:55

Have you properly understood why this theorem holds good?
– Arnab Chowdhury
Jul 29 at 9:12

Have you properly understood why this theorem holds good?
– Arnab Chowdhury
Jul 29 at 9:12

add a comment | 

2 Answers
2

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up vote
2
down vote



accepted

Take $1-x^2=1over1+p$,

$f_n(x) =n^2xover(1+p)^nleq n^2over(1+p)^n$ since $xin [0,1]$ Take $alpha=2$.

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answered Jul 29 at 9:19

Tsemo Aristide

50.9k11143

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With $$1-x^2=frac11+p$$ then $$x^2=fracp1+p$$ and while $p>0$ this concludes $x^2<1$ and then $0<x<1$. hence you can use the theorem by $alpha=2$. the case $x=1$ is trivial.

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edited Jul 29 at 11:05

answered Jul 29 at 9:39

Lolita

52318

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

Take $1-x^2=1over1+p$,

$f_n(x) =n^2xover(1+p)^nleq n^2over(1+p)^n$ since $xin [0,1]$ Take $alpha=2$.

share|cite|improve this answer

answered Jul 29 at 9:19

Tsemo Aristide

50.9k11143

add a comment | 

up vote
2
down vote



accepted

Take $1-x^2=1over1+p$,

$f_n(x) =n^2xover(1+p)^nleq n^2over(1+p)^n$ since $xin [0,1]$ Take $alpha=2$.

share|cite|improve this answer

answered Jul 29 at 9:19

Tsemo Aristide

50.9k11143

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

Take $1-x^2=1over1+p$,

$f_n(x) =n^2xover(1+p)^nleq n^2over(1+p)^n$ since $xin [0,1]$ Take $alpha=2$.

share|cite|improve this answer

answered Jul 29 at 9:19

Tsemo Aristide

50.9k11143

Take $1-x^2=1over1+p$,

$f_n(x) =n^2xover(1+p)^nleq n^2over(1+p)^n$ since $xin [0,1]$ Take $alpha=2$.

share|cite|improve this answer

answered Jul 29 at 9:19

Tsemo Aristide

50.9k11143

share|cite|improve this answer

share|cite|improve this answer

answered Jul 29 at 9:19

Tsemo Aristide

50.9k11143

answered Jul 29 at 9:19

Tsemo Aristide

50.9k11143

answered Jul 29 at 9:19

Tsemo Aristide

50.9k11143

50.9k11143

add a comment | 

add a comment | 

up vote
1
down vote

With $$1-x^2=frac11+p$$ then $$x^2=fracp1+p$$ and while $p>0$ this concludes $x^2<1$ and then $0<x<1$. hence you can use the theorem by $alpha=2$. the case $x=1$ is trivial.

share|cite|improve this answer

edited Jul 29 at 11:05

answered Jul 29 at 9:39

Lolita

52318

add a comment | 

up vote
1
down vote

With $$1-x^2=frac11+p$$ then $$x^2=fracp1+p$$ and while $p>0$ this concludes $x^2<1$ and then $0<x<1$. hence you can use the theorem by $alpha=2$. the case $x=1$ is trivial.

share|cite|improve this answer

edited Jul 29 at 11:05

answered Jul 29 at 9:39

Lolita

52318

add a comment | 

up vote
1
down vote

up vote
1
down vote

With $$1-x^2=frac11+p$$ then $$x^2=fracp1+p$$ and while $p>0$ this concludes $x^2<1$ and then $0<x<1$. hence you can use the theorem by $alpha=2$. the case $x=1$ is trivial.

share|cite|improve this answer

edited Jul 29 at 11:05

answered Jul 29 at 9:39

Lolita

52318

With $$1-x^2=frac11+p$$ then $$x^2=fracp1+p$$ and while $p>0$ this concludes $x^2<1$ and then $0<x<1$. hence you can use the theorem by $alpha=2$. the case $x=1$ is trivial.

share|cite|improve this answer

edited Jul 29 at 11:05

answered Jul 29 at 9:39

Lolita

52318

share|cite|improve this answer

share|cite|improve this answer

edited Jul 29 at 11:05

edited Jul 29 at 11:05

edited Jul 29 at 11:05

answered Jul 29 at 9:39

Lolita

52318

answered Jul 29 at 9:39

Lolita

52318

answered Jul 29 at 9:39

Lolita

52318

52318

add a comment | 

add a comment |