Proof verification: if $(x_n)to x$ and $forall ninmathbfN(x_nge 0)$, then $(sqrtx_n)tosqrtx$

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Is the following argument correct?

Proposition. If $(x_n)to x$ and $forall ninmathbfN(x_nge 0)$, show that $(sqrtx_n)tosqrtx$.

Proof. Assume $xneq 0$. Now since $x_nge 0,forall ninmathbfN$, appealing to theorem $textbf2.3.4$, yields $xge 0$ but $xneq 0$ and so $x>0$ and by extension $sqrtx>0$.

Let $epsilon>0$. Since $(x_n)to x$, there exists an $NinmathbfN$ such that $|x_n-x|<sqrtxepsilon$ whenever $nge N$. In addition since $x_nge 0$ it follows that $sqrtxge 0$ and thus $|sqrtx_n+sqrtx| = sqrtx_n+x$. Now consider the following observation.
$$|sqrtx_n-sqrtx| = fracsqrtx_n-sqrtx = fracsqrtx_n+sqrtxleq fracsqrtx$$
Now let $n$ be an arbitrary positive integer such that $nge N$, we have $|x_n-x|<sqrtxepsilon$, consequently
$|x_n-x|/sqrtx<epsilon$

$blacksquare$

Note:

• In the above proof the assumption of $xneq 0$ is due to the fact that i have proved the proposition in the event $x=0$.




• In addition theorem $textbf2.3.4$ is the statement that if $(x_n)to x$ then $forall ninmathbfN(x_nge 0)$ implies $xge 0$.

real-analysis proof-verification

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edited Jul 15 at 10:46

Alex Francisco

15.7k92047

asked Jul 15 at 10:08

Atif Farooq

2,7092824

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Some observations: (i) Second paragraph of proof, do you mean “whenever $n geq N$” not “$x geq N$”? (ii) Where do we see that $x_n geq 0$ for all $n$? (iii) I do not understand the wording in your bullet 2, it looks like shorthand for something that could be more easily written out clearly.
  – Michael
  Jul 15 at 10:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  My apologies let me make some corrections and yes i do mean $nge N$
  – Atif Farooq
  Jul 15 at 10:26
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Michael Bullet $2$ states that if $(x_n)$ converges to $x$ then if $x_nge 0$ for all $ninmathbfN$ then the limit $x$ must also be greater than equal to $0$.
  – Atif Farooq
  Jul 15 at 10:28
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I see. You fixed (i) and (ii) in my observations but for (ii) you added the text "$forall n in mathbfN(x_ngeq 0)$" in the statement of the proposition, which I do not understand but now I infer from your comment means "$x_n geq 0$ for all $n in mathbfN$." I would prefer the latter exposition as everyone can understand it. Anyway, except for this stylistic issue, everything looks fine. I just gave +1 for question.
  – Michael
  Jul 15 at 10:31
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Michael Thanks for your help and apologies for the inconvenience.
  – Atif Farooq
  Jul 15 at 10:36
  
  

  
  
  
  

add a comment | 

up vote
2
down vote

favorite

Is the following argument correct?

Proposition. If $(x_n)to x$ and $forall ninmathbfN(x_nge 0)$, show that $(sqrtx_n)tosqrtx$.

Proof. Assume $xneq 0$. Now since $x_nge 0,forall ninmathbfN$, appealing to theorem $textbf2.3.4$, yields $xge 0$ but $xneq 0$ and so $x>0$ and by extension $sqrtx>0$.

Let $epsilon>0$. Since $(x_n)to x$, there exists an $NinmathbfN$ such that $|x_n-x|<sqrtxepsilon$ whenever $nge N$. In addition since $x_nge 0$ it follows that $sqrtxge 0$ and thus $|sqrtx_n+sqrtx| = sqrtx_n+x$. Now consider the following observation.
$$|sqrtx_n-sqrtx| = fracsqrtx_n-sqrtx = fracsqrtx_n+sqrtxleq fracsqrtx$$
Now let $n$ be an arbitrary positive integer such that $nge N$, we have $|x_n-x|<sqrtxepsilon$, consequently
$|x_n-x|/sqrtx<epsilon$

$blacksquare$

Note:

• In the above proof the assumption of $xneq 0$ is due to the fact that i have proved the proposition in the event $x=0$.




• In addition theorem $textbf2.3.4$ is the statement that if $(x_n)to x$ then $forall ninmathbfN(x_nge 0)$ implies $xge 0$.

real-analysis proof-verification

share|cite|improve this question

edited Jul 15 at 10:46

Alex Francisco

15.7k92047

asked Jul 15 at 10:08

Atif Farooq

2,7092824

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Some observations: (i) Second paragraph of proof, do you mean “whenever $n geq N$” not “$x geq N$”? (ii) Where do we see that $x_n geq 0$ for all $n$? (iii) I do not understand the wording in your bullet 2, it looks like shorthand for something that could be more easily written out clearly.
  – Michael
  Jul 15 at 10:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  My apologies let me make some corrections and yes i do mean $nge N$
  – Atif Farooq
  Jul 15 at 10:26
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Michael Bullet $2$ states that if $(x_n)$ converges to $x$ then if $x_nge 0$ for all $ninmathbfN$ then the limit $x$ must also be greater than equal to $0$.
  – Atif Farooq
  Jul 15 at 10:28
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I see. You fixed (i) and (ii) in my observations but for (ii) you added the text "$forall n in mathbfN(x_ngeq 0)$" in the statement of the proposition, which I do not understand but now I infer from your comment means "$x_n geq 0$ for all $n in mathbfN$." I would prefer the latter exposition as everyone can understand it. Anyway, except for this stylistic issue, everything looks fine. I just gave +1 for question.
  – Michael
  Jul 15 at 10:31
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Michael Thanks for your help and apologies for the inconvenience.
  – Atif Farooq
  Jul 15 at 10:36
  
  

  
  
  
  

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

Is the following argument correct?

Proposition. If $(x_n)to x$ and $forall ninmathbfN(x_nge 0)$, show that $(sqrtx_n)tosqrtx$.

Proof. Assume $xneq 0$. Now since $x_nge 0,forall ninmathbfN$, appealing to theorem $textbf2.3.4$, yields $xge 0$ but $xneq 0$ and so $x>0$ and by extension $sqrtx>0$.

Let $epsilon>0$. Since $(x_n)to x$, there exists an $NinmathbfN$ such that $|x_n-x|<sqrtxepsilon$ whenever $nge N$. In addition since $x_nge 0$ it follows that $sqrtxge 0$ and thus $|sqrtx_n+sqrtx| = sqrtx_n+x$. Now consider the following observation.
$$|sqrtx_n-sqrtx| = fracsqrtx_n-sqrtx = fracsqrtx_n+sqrtxleq fracsqrtx$$
Now let $n$ be an arbitrary positive integer such that $nge N$, we have $|x_n-x|<sqrtxepsilon$, consequently
$|x_n-x|/sqrtx<epsilon$

$blacksquare$

Note:

• In the above proof the assumption of $xneq 0$ is due to the fact that i have proved the proposition in the event $x=0$.




• In addition theorem $textbf2.3.4$ is the statement that if $(x_n)to x$ then $forall ninmathbfN(x_nge 0)$ implies $xge 0$.

real-analysis proof-verification

share|cite|improve this question

edited Jul 15 at 10:46

Alex Francisco

15.7k92047

asked Jul 15 at 10:08

Atif Farooq

2,7092824

Is the following argument correct?

Proposition. If $(x_n)to x$ and $forall ninmathbfN(x_nge 0)$, show that $(sqrtx_n)tosqrtx$.

Proof. Assume $xneq 0$. Now since $x_nge 0,forall ninmathbfN$, appealing to theorem $textbf2.3.4$, yields $xge 0$ but $xneq 0$ and so $x>0$ and by extension $sqrtx>0$.

Let $epsilon>0$. Since $(x_n)to x$, there exists an $NinmathbfN$ such that $|x_n-x|<sqrtxepsilon$ whenever $nge N$. In addition since $x_nge 0$ it follows that $sqrtxge 0$ and thus $|sqrtx_n+sqrtx| = sqrtx_n+x$. Now consider the following observation.
$$|sqrtx_n-sqrtx| = fracsqrtx_n-sqrtx = fracsqrtx_n+sqrtxleq fracsqrtx$$
Now let $n$ be an arbitrary positive integer such that $nge N$, we have $|x_n-x|<sqrtxepsilon$, consequently
$|x_n-x|/sqrtx<epsilon$

$blacksquare$

Note:

• In the above proof the assumption of $xneq 0$ is due to the fact that i have proved the proposition in the event $x=0$.




• In addition theorem $textbf2.3.4$ is the statement that if $(x_n)to x$ then $forall ninmathbfN(x_nge 0)$ implies $xge 0$.

real-analysis proof-verification

share|cite|improve this question

edited Jul 15 at 10:46

Alex Francisco

15.7k92047

asked Jul 15 at 10:08

Atif Farooq

2,7092824

share|cite|improve this question

share|cite|improve this question

edited Jul 15 at 10:46

Alex Francisco

15.7k92047

edited Jul 15 at 10:46

Alex Francisco

15.7k92047

edited Jul 15 at 10:46

Alex Francisco

15.7k92047

15.7k92047

asked Jul 15 at 10:08

Atif Farooq

2,7092824

asked Jul 15 at 10:08

Atif Farooq

2,7092824

asked Jul 15 at 10:08

Atif Farooq

2,7092824

2,7092824

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Some observations: (i) Second paragraph of proof, do you mean “whenever $n geq N$” not “$x geq N$”? (ii) Where do we see that $x_n geq 0$ for all $n$? (iii) I do not understand the wording in your bullet 2, it looks like shorthand for something that could be more easily written out clearly.
  – Michael
  Jul 15 at 10:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  My apologies let me make some corrections and yes i do mean $nge N$
  – Atif Farooq
  Jul 15 at 10:26
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Michael Bullet $2$ states that if $(x_n)$ converges to $x$ then if $x_nge 0$ for all $ninmathbfN$ then the limit $x$ must also be greater than equal to $0$.
  – Atif Farooq
  Jul 15 at 10:28
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I see. You fixed (i) and (ii) in my observations but for (ii) you added the text "$forall n in mathbfN(x_ngeq 0)$" in the statement of the proposition, which I do not understand but now I infer from your comment means "$x_n geq 0$ for all $n in mathbfN$." I would prefer the latter exposition as everyone can understand it. Anyway, except for this stylistic issue, everything looks fine. I just gave +1 for question.
  – Michael
  Jul 15 at 10:31
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Michael Thanks for your help and apologies for the inconvenience.
  – Atif Farooq
  Jul 15 at 10:36
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Some observations: (i) Second paragraph of proof, do you mean “whenever $n geq N$” not “$x geq N$”? (ii) Where do we see that $x_n geq 0$ for all $n$? (iii) I do not understand the wording in your bullet 2, it looks like shorthand for something that could be more easily written out clearly.
  – Michael
  Jul 15 at 10:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  My apologies let me make some corrections and yes i do mean $nge N$
  – Atif Farooq
  Jul 15 at 10:26
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Michael Bullet $2$ states that if $(x_n)$ converges to $x$ then if $x_nge 0$ for all $ninmathbfN$ then the limit $x$ must also be greater than equal to $0$.
  – Atif Farooq
  Jul 15 at 10:28
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I see. You fixed (i) and (ii) in my observations but for (ii) you added the text "$forall n in mathbfN(x_ngeq 0)$" in the statement of the proposition, which I do not understand but now I infer from your comment means "$x_n geq 0$ for all $n in mathbfN$." I would prefer the latter exposition as everyone can understand it. Anyway, except for this stylistic issue, everything looks fine. I just gave +1 for question.
  – Michael
  Jul 15 at 10:31
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Michael Thanks for your help and apologies for the inconvenience.
  – Atif Farooq
  Jul 15 at 10:36
  
  

  
  
  
  

Some observations: (i) Second paragraph of proof, do you mean “whenever $n geq N$” not “$x geq N$”? (ii) Where do we see that $x_n geq 0$ for all $n$? (iii) I do not understand the wording in your bullet 2, it looks like shorthand for something that could be more easily written out clearly.
– Michael
Jul 15 at 10:21

Some observations: (i) Second paragraph of proof, do you mean “whenever $n geq N$” not “$x geq N$”? (ii) Where do we see that $x_n geq 0$ for all $n$? (iii) I do not understand the wording in your bullet 2, it looks like shorthand for something that could be more easily written out clearly.
– Michael
Jul 15 at 10:21

My apologies let me make some corrections and yes i do mean $nge N$
– Atif Farooq
Jul 15 at 10:26

My apologies let me make some corrections and yes i do mean $nge N$
– Atif Farooq
Jul 15 at 10:26

@Michael Bullet $2$ states that if $(x_n)$ converges to $x$ then if $x_nge 0$ for all $ninmathbfN$ then the limit $x$ must also be greater than equal to $0$.
– Atif Farooq
Jul 15 at 10:28

@Michael Bullet $2$ states that if $(x_n)$ converges to $x$ then if $x_nge 0$ for all $ninmathbfN$ then the limit $x$ must also be greater than equal to $0$.
– Atif Farooq
Jul 15 at 10:28

1

1

I see. You fixed (i) and (ii) in my observations but for (ii) you added the text "$forall n in mathbfN(x_ngeq 0)$" in the statement of the proposition, which I do not understand but now I infer from your comment means "$x_n geq 0$ for all $n in mathbfN$." I would prefer the latter exposition as everyone can understand it. Anyway, except for this stylistic issue, everything looks fine. I just gave +1 for question.
– Michael
Jul 15 at 10:31

I see. You fixed (i) and (ii) in my observations but for (ii) you added the text "$forall n in mathbfN(x_ngeq 0)$" in the statement of the proposition, which I do not understand but now I infer from your comment means "$x_n geq 0$ for all $n in mathbfN$." I would prefer the latter exposition as everyone can understand it. Anyway, except for this stylistic issue, everything looks fine. I just gave +1 for question.
– Michael
Jul 15 at 10:31

@Michael Thanks for your help and apologies for the inconvenience.
– Atif Farooq
Jul 15 at 10:36

@Michael Thanks for your help and apologies for the inconvenience.
– Atif Farooq
Jul 15 at 10:36

add a comment | 

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