Knowing the result of a given limit, calculate the following after. What is its result? [closed]

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So I know for sure that by resolving the following limit I will have that result.

$$ lim_xto 1fracsqrt[m]x-1x-1=frac1m$$

And now I have this:
$$ l=lim_xto 1frac(1-x)(1-sqrt[3]x)(1-sqrt[5]x)...(1-sqrt[n]x)(1-x)^n-1.$$
How can I resolve the last limit using the upper relation?

real-analysis analysis

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edited Jul 16 at 17:25

asked Jul 16 at 17:08

Robert Maracine

507

closed as off-topic by Lord Shark the Unknown, Shaun, Xander Henderson, Leucippus, Adrian Keister Jul 17 at 1:17

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." – Shaun, Xander Henderson, Leucippus, Adrian Keister

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

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How many things do you multiply in the nominator? $n-1$ And what is the power of denominator? $n-1$ also
  – Rumpelstiltskin
  Jul 16 at 17:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  it jumps from 1 to 3 then to 5 etc. So There are (n+1)/2 factors in nominator.
  – Robert Maracine
  Jul 16 at 17:11
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  No, square root of $2$, then square root of $3$, ..., square root of $n$. I understand the confusion, it's coming from the notational difference of square root of $2$
  – Rumpelstiltskin
  Jul 16 at 17:13
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What happened to the $1-sqrt[4]x$ factor?
  – Lord Shark the Unknown
  Jul 16 at 17:21
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @RobertMaracine But the first factor is $1-sqrt[2]x$.
  – Lord Shark the Unknown
  Jul 16 at 17:23
  

  
  
  
  

 | 
show 5 more comments

up vote
-2
down vote

favorite

So I know for sure that by resolving the following limit I will have that result.

$$ lim_xto 1fracsqrt[m]x-1x-1=frac1m$$

And now I have this:
$$ l=lim_xto 1frac(1-x)(1-sqrt[3]x)(1-sqrt[5]x)...(1-sqrt[n]x)(1-x)^n-1.$$
How can I resolve the last limit using the upper relation?

real-analysis analysis

share|cite|improve this question

edited Jul 16 at 17:25

asked Jul 16 at 17:08

Robert Maracine

507

closed as off-topic by Lord Shark the Unknown, Shaun, Xander Henderson, Leucippus, Adrian Keister Jul 17 at 1:17

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." – Shaun, Xander Henderson, Leucippus, Adrian Keister

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

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How many things do you multiply in the nominator? $n-1$ And what is the power of denominator? $n-1$ also
  – Rumpelstiltskin
  Jul 16 at 17:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  it jumps from 1 to 3 then to 5 etc. So There are (n+1)/2 factors in nominator.
  – Robert Maracine
  Jul 16 at 17:11
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  No, square root of $2$, then square root of $3$, ..., square root of $n$. I understand the confusion, it's coming from the notational difference of square root of $2$
  – Rumpelstiltskin
  Jul 16 at 17:13
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What happened to the $1-sqrt[4]x$ factor?
  – Lord Shark the Unknown
  Jul 16 at 17:21
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @RobertMaracine But the first factor is $1-sqrt[2]x$.
  – Lord Shark the Unknown
  Jul 16 at 17:23
  

  
  
  
  

 | 
show 5 more comments

up vote
-2
down vote

favorite

up vote
-2
down vote

favorite

So I know for sure that by resolving the following limit I will have that result.

$$ lim_xto 1fracsqrt[m]x-1x-1=frac1m$$

And now I have this:
$$ l=lim_xto 1frac(1-x)(1-sqrt[3]x)(1-sqrt[5]x)...(1-sqrt[n]x)(1-x)^n-1.$$
How can I resolve the last limit using the upper relation?

real-analysis analysis

share|cite|improve this question

edited Jul 16 at 17:25

asked Jul 16 at 17:08

Robert Maracine

507

So I know for sure that by resolving the following limit I will have that result.

$$ lim_xto 1fracsqrt[m]x-1x-1=frac1m$$

And now I have this:
$$ l=lim_xto 1frac(1-x)(1-sqrt[3]x)(1-sqrt[5]x)...(1-sqrt[n]x)(1-x)^n-1.$$
How can I resolve the last limit using the upper relation?

real-analysis analysis

share|cite|improve this question

edited Jul 16 at 17:25

asked Jul 16 at 17:08

Robert Maracine

507

share|cite|improve this question

share|cite|improve this question

edited Jul 16 at 17:25

edited Jul 16 at 17:25

edited Jul 16 at 17:25

asked Jul 16 at 17:08

Robert Maracine

507

asked Jul 16 at 17:08

Robert Maracine

507

asked Jul 16 at 17:08

Robert Maracine

507

507

closed as off-topic by Lord Shark the Unknown, Shaun, Xander Henderson, Leucippus, Adrian Keister Jul 17 at 1:17

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." – Shaun, Xander Henderson, Leucippus, Adrian Keister

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

closed as off-topic by Lord Shark the Unknown, Shaun, Xander Henderson, Leucippus, Adrian Keister Jul 17 at 1:17

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." – Shaun, Xander Henderson, Leucippus, Adrian Keister

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

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How many things do you multiply in the nominator? $n-1$ And what is the power of denominator? $n-1$ also
  – Rumpelstiltskin
  Jul 16 at 17:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  it jumps from 1 to 3 then to 5 etc. So There are (n+1)/2 factors in nominator.
  – Robert Maracine
  Jul 16 at 17:11
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  No, square root of $2$, then square root of $3$, ..., square root of $n$. I understand the confusion, it's coming from the notational difference of square root of $2$
  – Rumpelstiltskin
  Jul 16 at 17:13
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What happened to the $1-sqrt[4]x$ factor?
  – Lord Shark the Unknown
  Jul 16 at 17:21
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @RobertMaracine But the first factor is $1-sqrt[2]x$.
  – Lord Shark the Unknown
  Jul 16 at 17:23
  

  
  
  
  

 | 
show 5 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How many things do you multiply in the nominator? $n-1$ And what is the power of denominator? $n-1$ also
  – Rumpelstiltskin
  Jul 16 at 17:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  it jumps from 1 to 3 then to 5 etc. So There are (n+1)/2 factors in nominator.
  – Robert Maracine
  Jul 16 at 17:11
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  No, square root of $2$, then square root of $3$, ..., square root of $n$. I understand the confusion, it's coming from the notational difference of square root of $2$
  – Rumpelstiltskin
  Jul 16 at 17:13
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What happened to the $1-sqrt[4]x$ factor?
  – Lord Shark the Unknown
  Jul 16 at 17:21
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @RobertMaracine But the first factor is $1-sqrt[2]x$.
  – Lord Shark the Unknown
  Jul 16 at 17:23
  

  
  
  
  

How many things do you multiply in the nominator? $n-1$ And what is the power of denominator? $n-1$ also
– Rumpelstiltskin
Jul 16 at 17:10

How many things do you multiply in the nominator? $n-1$ And what is the power of denominator? $n-1$ also
– Rumpelstiltskin
Jul 16 at 17:10

it jumps from 1 to 3 then to 5 etc. So There are (n+1)/2 factors in nominator.
– Robert Maracine
Jul 16 at 17:11

it jumps from 1 to 3 then to 5 etc. So There are (n+1)/2 factors in nominator.
– Robert Maracine
Jul 16 at 17:11

No, square root of $2$, then square root of $3$, ..., square root of $n$. I understand the confusion, it's coming from the notational difference of square root of $2$
– Rumpelstiltskin
Jul 16 at 17:13

No, square root of $2$, then square root of $3$, ..., square root of $n$. I understand the confusion, it's coming from the notational difference of square root of $2$
– Rumpelstiltskin
Jul 16 at 17:13

What happened to the $1-sqrt[4]x$ factor?
– Lord Shark the Unknown
Jul 16 at 17:21

What happened to the $1-sqrt[4]x$ factor?
– Lord Shark the Unknown
Jul 16 at 17:21

1

1

@RobertMaracine But the first factor is $1-sqrt[2]x$.
– Lord Shark the Unknown
Jul 16 at 17:23

@RobertMaracine But the first factor is $1-sqrt[2]x$.
– Lord Shark the Unknown
Jul 16 at 17:23

 | 
show 5 more comments

2 Answers
2

active

oldest

votes

up vote
2
down vote



accepted

Your limit is equal to

$$ lim_xto1 frac1-sqrtx1-x cdot lim_xto1 frac1-sqrt[3]x1-xcdotcdotscdotlim_xto1frac1-sqrt[n]x1-x cdot lim_xto 1 frac1(1-x)^n-1-leftlceilfracn2rightrceil = 1cdotfrac13 cdot frac15cdotsfrac1n cdot +infty = +infty$$

share|cite|improve this answer

answered Jul 16 at 17:15

mechanodroid

22.3k52041

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Where does$$frac1(1-x)^n-1-lceilfrac n2rceil$$come from?
  – Lord Shark the Unknown
  Jul 16 at 17:17
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  yes, i did that too, but the examinator who gave me the problem said that the result is 1/n! .
  – Robert Maracine
  Jul 16 at 17:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @LordSharktheUnknown There are $lceilfrac n2rceil$ terms in the numerator so I factored out $lceilfrac n2rceil$ copies of $1-x$.
  – mechanodroid
  Jul 16 at 17:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @mechanodroid I saw $n-1$ terms in the (original) version of the question. But I haven't a clue how many there are now $ddotfrown$.
  – Lord Shark the Unknown
  Jul 16 at 17:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @RobertMaracine the result can not be $frac1n!$
  – Ahmad Bazzi
  Jul 16 at 17:27
  

  
  
  
  

 | 
show 1 more comment

up vote
2
down vote

By the product rule,
$$l=prod_k=2^nlim_xto 1left(fracsqrt[k]x-1x-1right)
=prod_k=2^ncdots.$$

share|cite|improve this answer

answered Jul 16 at 17:15

Lord Shark the Unknown

85.6k951112

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Numerator contains odd roots $1,3,5 ldots n$
  – Ahmad Bazzi
  Jul 16 at 17:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @AhmadBazzi No, it doesn't.
  – Lord Shark the Unknown
  Jul 16 at 17:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  yes it does.., let me edit that so it's obvious
  – Robert Maracine
  Jul 16 at 17:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @LordSharktheUnknown thank you for the advice and sorry for the misunderstanding!
  – Robert Maracine
  Jul 16 at 17:32
  

  
  
  
  

add a comment | 

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

Your limit is equal to

$$ lim_xto1 frac1-sqrtx1-x cdot lim_xto1 frac1-sqrt[3]x1-xcdotcdotscdotlim_xto1frac1-sqrt[n]x1-x cdot lim_xto 1 frac1(1-x)^n-1-leftlceilfracn2rightrceil = 1cdotfrac13 cdot frac15cdotsfrac1n cdot +infty = +infty$$

share|cite|improve this answer

answered Jul 16 at 17:15

mechanodroid

22.3k52041

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Where does$$frac1(1-x)^n-1-lceilfrac n2rceil$$come from?
  – Lord Shark the Unknown
  Jul 16 at 17:17
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  yes, i did that too, but the examinator who gave me the problem said that the result is 1/n! .
  – Robert Maracine
  Jul 16 at 17:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @LordSharktheUnknown There are $lceilfrac n2rceil$ terms in the numerator so I factored out $lceilfrac n2rceil$ copies of $1-x$.
  – mechanodroid
  Jul 16 at 17:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @mechanodroid I saw $n-1$ terms in the (original) version of the question. But I haven't a clue how many there are now $ddotfrown$.
  – Lord Shark the Unknown
  Jul 16 at 17:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @RobertMaracine the result can not be $frac1n!$
  – Ahmad Bazzi
  Jul 16 at 17:27
  

  
  
  
  

 | 
show 1 more comment

up vote
2
down vote



accepted

Your limit is equal to

$$ lim_xto1 frac1-sqrtx1-x cdot lim_xto1 frac1-sqrt[3]x1-xcdotcdotscdotlim_xto1frac1-sqrt[n]x1-x cdot lim_xto 1 frac1(1-x)^n-1-leftlceilfracn2rightrceil = 1cdotfrac13 cdot frac15cdotsfrac1n cdot +infty = +infty$$

share|cite|improve this answer

answered Jul 16 at 17:15

mechanodroid

22.3k52041

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Where does$$frac1(1-x)^n-1-lceilfrac n2rceil$$come from?
  – Lord Shark the Unknown
  Jul 16 at 17:17
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  yes, i did that too, but the examinator who gave me the problem said that the result is 1/n! .
  – Robert Maracine
  Jul 16 at 17:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @LordSharktheUnknown There are $lceilfrac n2rceil$ terms in the numerator so I factored out $lceilfrac n2rceil$ copies of $1-x$.
  – mechanodroid
  Jul 16 at 17:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @mechanodroid I saw $n-1$ terms in the (original) version of the question. But I haven't a clue how many there are now $ddotfrown$.
  – Lord Shark the Unknown
  Jul 16 at 17:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @RobertMaracine the result can not be $frac1n!$
  – Ahmad Bazzi
  Jul 16 at 17:27
  

  
  
  
  

 | 
show 1 more comment

up vote
2
down vote



accepted

up vote
2
down vote



accepted

Your limit is equal to

$$ lim_xto1 frac1-sqrtx1-x cdot lim_xto1 frac1-sqrt[3]x1-xcdotcdotscdotlim_xto1frac1-sqrt[n]x1-x cdot lim_xto 1 frac1(1-x)^n-1-leftlceilfracn2rightrceil = 1cdotfrac13 cdot frac15cdotsfrac1n cdot +infty = +infty$$

share|cite|improve this answer

answered Jul 16 at 17:15

mechanodroid

22.3k52041

Your limit is equal to

$$ lim_xto1 frac1-sqrtx1-x cdot lim_xto1 frac1-sqrt[3]x1-xcdotcdotscdotlim_xto1frac1-sqrt[n]x1-x cdot lim_xto 1 frac1(1-x)^n-1-leftlceilfracn2rightrceil = 1cdotfrac13 cdot frac15cdotsfrac1n cdot +infty = +infty$$

share|cite|improve this answer

answered Jul 16 at 17:15

mechanodroid

22.3k52041

share|cite|improve this answer

share|cite|improve this answer

answered Jul 16 at 17:15

mechanodroid

22.3k52041

answered Jul 16 at 17:15

mechanodroid

22.3k52041

answered Jul 16 at 17:15

mechanodroid

22.3k52041

22.3k52041

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Where does$$frac1(1-x)^n-1-lceilfrac n2rceil$$come from?
  – Lord Shark the Unknown
  Jul 16 at 17:17
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  yes, i did that too, but the examinator who gave me the problem said that the result is 1/n! .
  – Robert Maracine
  Jul 16 at 17:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @LordSharktheUnknown There are $lceilfrac n2rceil$ terms in the numerator so I factored out $lceilfrac n2rceil$ copies of $1-x$.
  – mechanodroid
  Jul 16 at 17:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @mechanodroid I saw $n-1$ terms in the (original) version of the question. But I haven't a clue how many there are now $ddotfrown$.
  – Lord Shark the Unknown
  Jul 16 at 17:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @RobertMaracine the result can not be $frac1n!$
  – Ahmad Bazzi
  Jul 16 at 17:27
  

  
  
  
  

 | 
show 1 more comment

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Where does$$frac1(1-x)^n-1-lceilfrac n2rceil$$come from?
  – Lord Shark the Unknown
  Jul 16 at 17:17
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  yes, i did that too, but the examinator who gave me the problem said that the result is 1/n! .
  – Robert Maracine
  Jul 16 at 17:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @LordSharktheUnknown There are $lceilfrac n2rceil$ terms in the numerator so I factored out $lceilfrac n2rceil$ copies of $1-x$.
  – mechanodroid
  Jul 16 at 17:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @mechanodroid I saw $n-1$ terms in the (original) version of the question. But I haven't a clue how many there are now $ddotfrown$.
  – Lord Shark the Unknown
  Jul 16 at 17:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @RobertMaracine the result can not be $frac1n!$
  – Ahmad Bazzi
  Jul 16 at 17:27
  

  
  
  
  

1

1

Where does$$frac1(1-x)^n-1-lceilfrac n2rceil$$come from?
– Lord Shark the Unknown
Jul 16 at 17:17

Where does$$frac1(1-x)^n-1-lceilfrac n2rceil$$come from?
– Lord Shark the Unknown
Jul 16 at 17:17

yes, i did that too, but the examinator who gave me the problem said that the result is 1/n! .
– Robert Maracine
Jul 16 at 17:17

yes, i did that too, but the examinator who gave me the problem said that the result is 1/n! .
– Robert Maracine
Jul 16 at 17:17

@LordSharktheUnknown There are $lceilfrac n2rceil$ terms in the numerator so I factored out $lceilfrac n2rceil$ copies of $1-x$.
– mechanodroid
Jul 16 at 17:23

@LordSharktheUnknown There are $lceilfrac n2rceil$ terms in the numerator so I factored out $lceilfrac n2rceil$ copies of $1-x$.
– mechanodroid
Jul 16 at 17:23

@mechanodroid I saw $n-1$ terms in the (original) version of the question. But I haven't a clue how many there are now $ddotfrown$.
– Lord Shark the Unknown
Jul 16 at 17:24

@mechanodroid I saw $n-1$ terms in the (original) version of the question. But I haven't a clue how many there are now $ddotfrown$.
– Lord Shark the Unknown
Jul 16 at 17:24

@RobertMaracine the result can not be $frac1n!$
– Ahmad Bazzi
Jul 16 at 17:27

@RobertMaracine the result can not be $frac1n!$
– Ahmad Bazzi
Jul 16 at 17:27

 | 
show 1 more comment

up vote
2
down vote

By the product rule,
$$l=prod_k=2^nlim_xto 1left(fracsqrt[k]x-1x-1right)
=prod_k=2^ncdots.$$

share|cite|improve this answer

answered Jul 16 at 17:15

Lord Shark the Unknown

85.6k951112

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Numerator contains odd roots $1,3,5 ldots n$
  – Ahmad Bazzi
  Jul 16 at 17:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @AhmadBazzi No, it doesn't.
  – Lord Shark the Unknown
  Jul 16 at 17:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  yes it does.., let me edit that so it's obvious
  – Robert Maracine
  Jul 16 at 17:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @LordSharktheUnknown thank you for the advice and sorry for the misunderstanding!
  – Robert Maracine
  Jul 16 at 17:32
  

  
  
  
  

add a comment | 

up vote
2
down vote

By the product rule,
$$l=prod_k=2^nlim_xto 1left(fracsqrt[k]x-1x-1right)
=prod_k=2^ncdots.$$

share|cite|improve this answer

answered Jul 16 at 17:15

Lord Shark the Unknown

85.6k951112

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Numerator contains odd roots $1,3,5 ldots n$
  – Ahmad Bazzi
  Jul 16 at 17:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @AhmadBazzi No, it doesn't.
  – Lord Shark the Unknown
  Jul 16 at 17:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  yes it does.., let me edit that so it's obvious
  – Robert Maracine
  Jul 16 at 17:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @LordSharktheUnknown thank you for the advice and sorry for the misunderstanding!
  – Robert Maracine
  Jul 16 at 17:32
  

  
  
  
  

add a comment | 

up vote
2
down vote

up vote
2
down vote

By the product rule,
$$l=prod_k=2^nlim_xto 1left(fracsqrt[k]x-1x-1right)
=prod_k=2^ncdots.$$

share|cite|improve this answer

answered Jul 16 at 17:15

Lord Shark the Unknown

85.6k951112

By the product rule,
$$l=prod_k=2^nlim_xto 1left(fracsqrt[k]x-1x-1right)
=prod_k=2^ncdots.$$

share|cite|improve this answer

answered Jul 16 at 17:15

Lord Shark the Unknown

85.6k951112

share|cite|improve this answer

share|cite|improve this answer

answered Jul 16 at 17:15

Lord Shark the Unknown

85.6k951112

answered Jul 16 at 17:15

Lord Shark the Unknown

85.6k951112

answered Jul 16 at 17:15

Lord Shark the Unknown

85.6k951112

85.6k951112

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Numerator contains odd roots $1,3,5 ldots n$
  – Ahmad Bazzi
  Jul 16 at 17:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @AhmadBazzi No, it doesn't.
  – Lord Shark the Unknown
  Jul 16 at 17:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  yes it does.., let me edit that so it's obvious
  – Robert Maracine
  Jul 16 at 17:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @LordSharktheUnknown thank you for the advice and sorry for the misunderstanding!
  – Robert Maracine
  Jul 16 at 17:32
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Numerator contains odd roots $1,3,5 ldots n$
  – Ahmad Bazzi
  Jul 16 at 17:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @AhmadBazzi No, it doesn't.
  – Lord Shark the Unknown
  Jul 16 at 17:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  yes it does.., let me edit that so it's obvious
  – Robert Maracine
  Jul 16 at 17:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @LordSharktheUnknown thank you for the advice and sorry for the misunderstanding!
  – Robert Maracine
  Jul 16 at 17:32
  

  
  
  
  

Numerator contains odd roots $1,3,5 ldots n$
– Ahmad Bazzi
Jul 16 at 17:18

Numerator contains odd roots $1,3,5 ldots n$
– Ahmad Bazzi
Jul 16 at 17:18

@AhmadBazzi No, it doesn't.
– Lord Shark the Unknown
Jul 16 at 17:19

@AhmadBazzi No, it doesn't.
– Lord Shark the Unknown
Jul 16 at 17:19

yes it does.., let me edit that so it's obvious
– Robert Maracine
Jul 16 at 17:19

yes it does.., let me edit that so it's obvious
– Robert Maracine
Jul 16 at 17:19

@LordSharktheUnknown thank you for the advice and sorry for the misunderstanding!
– Robert Maracine
Jul 16 at 17:32

@LordSharktheUnknown thank you for the advice and sorry for the misunderstanding!
– Robert Maracine
Jul 16 at 17:32

add a comment |