Problems proving that $limlimits_{n to infty}frac{2n}{n^3+1}=0 $ [closed]

-1

I have to prove using the definition of a limit.

Following the definition I think I should find n for which it holds: $lvertfrac{2n}{n^3+1}rvertltepsilon$

But after some transformations I end up with: $frac{2}{epsilon}lt n^2+frac{1}{n}$

and I don't know where to go from now on

limits limits-without-lhopital

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asked Nov 20 '18 at 16:50

Bartosz

315

closed as off-topic by Nosrati, amWhy, Shailesh, user10354138, Cesareo Nov 21 '18 at 2:13

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." – Nosrati, amWhy, Shailesh, user10354138, Cesareo

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

• 
  
  
  

  
  

  
  
  
  
  
  

  
  If $n^2>2epsilon$, then $n^2+1/n>2epsilon$ as well.
  – Yves Daoust
  Nov 20 '18 at 17:03
  
  
  

  
  
  
  

add a comment | 

-1

I have to prove using the definition of a limit.

Following the definition I think I should find n for which it holds: $lvertfrac{2n}{n^3+1}rvertltepsilon$

But after some transformations I end up with: $frac{2}{epsilon}lt n^2+frac{1}{n}$

and I don't know where to go from now on

limits limits-without-lhopital

share|cite|improve this question

asked Nov 20 '18 at 16:50

Bartosz

315

closed as off-topic by Nosrati, amWhy, Shailesh, user10354138, Cesareo Nov 21 '18 at 2:13

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." – Nosrati, amWhy, Shailesh, user10354138, Cesareo

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

• 
  
  
  

  
  

  
  
  
  
  
  

  
  If $n^2>2epsilon$, then $n^2+1/n>2epsilon$ as well.
  – Yves Daoust
  Nov 20 '18 at 17:03
  
  
  

  
  
  
  

add a comment | 

-1

-1

-1

I have to prove using the definition of a limit.

Following the definition I think I should find n for which it holds: $lvertfrac{2n}{n^3+1}rvertltepsilon$

But after some transformations I end up with: $frac{2}{epsilon}lt n^2+frac{1}{n}$

and I don't know where to go from now on

limits limits-without-lhopital

share|cite|improve this question

asked Nov 20 '18 at 16:50

Bartosz

315

I have to prove using the definition of a limit.

Following the definition I think I should find n for which it holds: $lvertfrac{2n}{n^3+1}rvertltepsilon$

But after some transformations I end up with: $frac{2}{epsilon}lt n^2+frac{1}{n}$

and I don't know where to go from now on

limits limits-without-lhopital

limits limits-without-lhopital

share|cite|improve this question

asked Nov 20 '18 at 16:50

Bartosz

315

share|cite|improve this question

asked Nov 20 '18 at 16:50

Bartosz

315

share|cite|improve this question

share|cite|improve this question

asked Nov 20 '18 at 16:50

Bartosz

315

asked Nov 20 '18 at 16:50

Bartosz

315

asked Nov 20 '18 at 16:50

Bartosz

315

315

closed as off-topic by Nosrati, amWhy, Shailesh, user10354138, Cesareo Nov 21 '18 at 2:13

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." – Nosrati, amWhy, Shailesh, user10354138, Cesareo

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

closed as off-topic by Nosrati, amWhy, Shailesh, user10354138, Cesareo Nov 21 '18 at 2:13

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." – Nosrati, amWhy, Shailesh, user10354138, Cesareo

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

• 
  
  
  

  
  

  
  
  
  
  
  

  
  If $n^2>2epsilon$, then $n^2+1/n>2epsilon$ as well.
  – Yves Daoust
  Nov 20 '18 at 17:03
  
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  If $n^2>2epsilon$, then $n^2+1/n>2epsilon$ as well.
  – Yves Daoust
  Nov 20 '18 at 17:03
  
  
  

  
  
  
  

If $n^2>2epsilon$, then $n^2+1/n>2epsilon$ as well.
– Yves Daoust
Nov 20 '18 at 17:03

If $n^2>2epsilon$, then $n^2+1/n>2epsilon$ as well.
– Yves Daoust
Nov 20 '18 at 17:03

add a comment | 

5 Answers
5

active

oldest

votes

1

HINT: Note that $${2nover n^3 + 1} le {2nover n^3} = {2over n^2}.$$

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answered Nov 20 '18 at 16:53

Decaf-Math

3,209825

add a comment | 

1

Note that
$$
frac{2n}{n^3+1}leq frac{2n}{n^3}=frac{2}{n^2}<epsilon
$$
if $n>sqrt{2/epsilon}$.

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answered Nov 20 '18 at 16:53

Foobaz John

21.3k41251

add a comment | 

1

Hint

$dfrac{2n}{n^3+1}leqdfrac{2n}{n^3}=dfrac{2}{n^2}$

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answered Nov 20 '18 at 16:54

giannispapav

1,534324

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0

Hint:
$$dfrac{2n}{n^3+1}=dfrac{2}{(n+1)(n+frac1n-1)}$$
also
$$n+frac1ngeq2$$

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answered Nov 20 '18 at 17:02

Nosrati

26.5k62353

add a comment | 

0

We have that

$$left|frac{2n}{n^3+1}right|=frac{2n}{n^3+1}ltepsilon iff epsilon n^3-2n+epsilon>0$$

and for $n$ large

$$epsilon n^3-2n+epsilon>epsilon n^2-2n=n(epsilon n-2)>0 iff n>frac 2 epsilon $$

share|cite|improve this answer

edited Nov 20 '18 at 17:15

answered Nov 20 '18 at 17:10

gimusi

1

add a comment | 

5 Answers
5

active

oldest

votes

5 Answers
5

active

oldest

votes

active

oldest

votes

active

oldest

votes

1

HINT: Note that $${2nover n^3 + 1} le {2nover n^3} = {2over n^2}.$$

share|cite|improve this answer

answered Nov 20 '18 at 16:53

Decaf-Math

3,209825

add a comment | 

1

HINT: Note that $${2nover n^3 + 1} le {2nover n^3} = {2over n^2}.$$

share|cite|improve this answer

answered Nov 20 '18 at 16:53

Decaf-Math

3,209825

add a comment | 

1

1

1

HINT: Note that $${2nover n^3 + 1} le {2nover n^3} = {2over n^2}.$$

share|cite|improve this answer

answered Nov 20 '18 at 16:53

Decaf-Math

3,209825

HINT: Note that $${2nover n^3 + 1} le {2nover n^3} = {2over n^2}.$$

share|cite|improve this answer

answered Nov 20 '18 at 16:53

Decaf-Math

3,209825

share|cite|improve this answer

share|cite|improve this answer

answered Nov 20 '18 at 16:53

Decaf-Math

3,209825

answered Nov 20 '18 at 16:53

Decaf-Math

3,209825

answered Nov 20 '18 at 16:53

Decaf-Math

3,209825

3,209825

add a comment | 

add a comment | 

1

Note that
$$
frac{2n}{n^3+1}leq frac{2n}{n^3}=frac{2}{n^2}<epsilon
$$
if $n>sqrt{2/epsilon}$.

share|cite|improve this answer

answered Nov 20 '18 at 16:53

Foobaz John

21.3k41251

add a comment | 

1

Note that
$$
frac{2n}{n^3+1}leq frac{2n}{n^3}=frac{2}{n^2}<epsilon
$$
if $n>sqrt{2/epsilon}$.

share|cite|improve this answer

answered Nov 20 '18 at 16:53

Foobaz John

21.3k41251

add a comment | 

1

1

1

Note that
$$
frac{2n}{n^3+1}leq frac{2n}{n^3}=frac{2}{n^2}<epsilon
$$
if $n>sqrt{2/epsilon}$.

share|cite|improve this answer

answered Nov 20 '18 at 16:53

Foobaz John

21.3k41251

Note that
$$
frac{2n}{n^3+1}leq frac{2n}{n^3}=frac{2}{n^2}<epsilon
$$
if $n>sqrt{2/epsilon}$.

share|cite|improve this answer

answered Nov 20 '18 at 16:53

Foobaz John

21.3k41251

share|cite|improve this answer

share|cite|improve this answer

answered Nov 20 '18 at 16:53

Foobaz John

21.3k41251

answered Nov 20 '18 at 16:53

Foobaz John

21.3k41251

answered Nov 20 '18 at 16:53

Foobaz John

21.3k41251

21.3k41251

add a comment | 

add a comment | 

1

Hint

$dfrac{2n}{n^3+1}leqdfrac{2n}{n^3}=dfrac{2}{n^2}$

share|cite|improve this answer

answered Nov 20 '18 at 16:54

giannispapav

1,534324

add a comment | 

1

Hint

$dfrac{2n}{n^3+1}leqdfrac{2n}{n^3}=dfrac{2}{n^2}$

share|cite|improve this answer

answered Nov 20 '18 at 16:54

giannispapav

1,534324

add a comment | 

1

1

1

Hint

$dfrac{2n}{n^3+1}leqdfrac{2n}{n^3}=dfrac{2}{n^2}$

share|cite|improve this answer

answered Nov 20 '18 at 16:54

giannispapav

1,534324

Hint

$dfrac{2n}{n^3+1}leqdfrac{2n}{n^3}=dfrac{2}{n^2}$

share|cite|improve this answer

answered Nov 20 '18 at 16:54

giannispapav

1,534324

share|cite|improve this answer

share|cite|improve this answer

answered Nov 20 '18 at 16:54

giannispapav

1,534324

answered Nov 20 '18 at 16:54

giannispapav

1,534324

answered Nov 20 '18 at 16:54

giannispapav

1,534324

1,534324

add a comment | 

add a comment | 

0

Hint:
$$dfrac{2n}{n^3+1}=dfrac{2}{(n+1)(n+frac1n-1)}$$
also
$$n+frac1ngeq2$$

share|cite|improve this answer

answered Nov 20 '18 at 17:02

Nosrati

26.5k62353

add a comment | 

0

Hint:
$$dfrac{2n}{n^3+1}=dfrac{2}{(n+1)(n+frac1n-1)}$$
also
$$n+frac1ngeq2$$

share|cite|improve this answer

answered Nov 20 '18 at 17:02

Nosrati

26.5k62353

add a comment | 

0

0

0

Hint:
$$dfrac{2n}{n^3+1}=dfrac{2}{(n+1)(n+frac1n-1)}$$
also
$$n+frac1ngeq2$$

share|cite|improve this answer

answered Nov 20 '18 at 17:02

Nosrati

26.5k62353

Hint:
$$dfrac{2n}{n^3+1}=dfrac{2}{(n+1)(n+frac1n-1)}$$
also
$$n+frac1ngeq2$$

share|cite|improve this answer

answered Nov 20 '18 at 17:02

Nosrati

26.5k62353

share|cite|improve this answer

share|cite|improve this answer

answered Nov 20 '18 at 17:02

Nosrati

26.5k62353

answered Nov 20 '18 at 17:02

Nosrati

26.5k62353

answered Nov 20 '18 at 17:02

Nosrati

26.5k62353

26.5k62353

add a comment | 

add a comment | 

0

We have that

$$left|frac{2n}{n^3+1}right|=frac{2n}{n^3+1}ltepsilon iff epsilon n^3-2n+epsilon>0$$

and for $n$ large

$$epsilon n^3-2n+epsilon>epsilon n^2-2n=n(epsilon n-2)>0 iff n>frac 2 epsilon $$

share|cite|improve this answer

edited Nov 20 '18 at 17:15

answered Nov 20 '18 at 17:10

gimusi

1

add a comment | 

0

We have that

$$left|frac{2n}{n^3+1}right|=frac{2n}{n^3+1}ltepsilon iff epsilon n^3-2n+epsilon>0$$

and for $n$ large

$$epsilon n^3-2n+epsilon>epsilon n^2-2n=n(epsilon n-2)>0 iff n>frac 2 epsilon $$

share|cite|improve this answer

edited Nov 20 '18 at 17:15

answered Nov 20 '18 at 17:10

gimusi

1

add a comment | 

0

0

0

We have that

$$left|frac{2n}{n^3+1}right|=frac{2n}{n^3+1}ltepsilon iff epsilon n^3-2n+epsilon>0$$

and for $n$ large

$$epsilon n^3-2n+epsilon>epsilon n^2-2n=n(epsilon n-2)>0 iff n>frac 2 epsilon $$

share|cite|improve this answer

edited Nov 20 '18 at 17:15

answered Nov 20 '18 at 17:10

gimusi

1

We have that

$$left|frac{2n}{n^3+1}right|=frac{2n}{n^3+1}ltepsilon iff epsilon n^3-2n+epsilon>0$$

and for $n$ large

$$epsilon n^3-2n+epsilon>epsilon n^2-2n=n(epsilon n-2)>0 iff n>frac 2 epsilon $$

share|cite|improve this answer

edited Nov 20 '18 at 17:15

answered Nov 20 '18 at 17:10

gimusi

1

share|cite|improve this answer

share|cite|improve this answer

edited Nov 20 '18 at 17:15

edited Nov 20 '18 at 17:15

edited Nov 20 '18 at 17:15

answered Nov 20 '18 at 17:10

gimusi

1

answered Nov 20 '18 at 17:10

gimusi

1

answered Nov 20 '18 at 17:10

gimusi

1

1

add a comment | 

add a comment | 

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