Proving $lim_{ntoinfty}sqrt{n^2+n}-n=frac{1}{2}$












2












$begingroup$



Prove using the definition of a limit that
$$displaystyle{lim_{ntoinfty}underbrace{sqrt{n^2+n}-n}_{s_n}}=frac{1}{2}.$$




Proof:



Let $epsilon>0$ and $ninmathbb{N}$, then
$$left|sqrt{n^2+n}-n-frac12right|stackrel{(*)}{=}left|frac{n-sqrt{n^2+n}}{2(n+sqrt{n^2+n})}right|=left|frac{n}{2(n+sqrt{n^2+n})^2}right|=left|frac{n}{4n^2+4nsqrt{n^2+n}+2n}right|<left|frac{n}{4nsqrt{n^2+n}+2n}right|=left|frac{1}{4sqrt{n^2+n}+2}right|$$
Since $forall ninmathbb{N},,sqrt{n^2+n}>sqrt{n},$ then
$$left|frac{1}{4sqrt{n^2+n}+2}right|<left|frac{1}{4sqrt{n}}right|<epsilonquadtext{ if }quad n>frac{1}{16epsilon^2}.$$



Would this be correct?




(*) maybe a simpler way would be to recognize that the numerator is $|s_n|<1?$






real-analysis sequences-and-series limits proof-verification







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$endgroup$












  • $begingroup$
    Abit lost at your first expression after =, why don’t you just multiply the original by an expression to complete the difference of squares (and divide, respectively, to compensate), and then just divide numerator and denominator by $n$?
    $endgroup$
    – Makina
    Dec 15 '18 at 5:49








  • 2




    $begingroup$
    A more direct way is to note that $$sqrt{n^2+n}-n=frac1{sqrt{1+frac1n}+1}$$ whose limit is direct, as has been explained tons of times on the site (did you search for duplicates before posting this?).
    $endgroup$
    – Did
    Dec 15 '18 at 5:50










  • $begingroup$
    @Did not sure where that leads me to... I am supposed to use the definition of a limit to prove the statement, not just compute it. I am able to compute the limit, but that's not the scope here
    $endgroup$
    – DMH16
    Dec 15 '18 at 5:53












  • $begingroup$
    @Makina that's what I did, but then dividing by $n$ as mentioned by Did does not really help a whole lot, since I am not trying to compute the limit with limit properties, but show that the equality holds using the definition
    $endgroup$
    – DMH16
    Dec 15 '18 at 5:54








  • 1




    $begingroup$
    @DMH16 Your algebra and logic are correct, what part of the proof are you uncertain about? Can you justify the use of inequalities to yourself?
    $endgroup$
    – DanielV
    Dec 15 '18 at 11:13
















2












$begingroup$



Prove using the definition of a limit that
$$displaystyle{lim_{ntoinfty}underbrace{sqrt{n^2+n}-n}_{s_n}}=frac{1}{2}.$$




Proof:



Let $epsilon>0$ and $ninmathbb{N}$, then
$$left|sqrt{n^2+n}-n-frac12right|stackrel{(*)}{=}left|frac{n-sqrt{n^2+n}}{2(n+sqrt{n^2+n})}right|=left|frac{n}{2(n+sqrt{n^2+n})^2}right|=left|frac{n}{4n^2+4nsqrt{n^2+n}+2n}right|<left|frac{n}{4nsqrt{n^2+n}+2n}right|=left|frac{1}{4sqrt{n^2+n}+2}right|$$
Since $forall ninmathbb{N},,sqrt{n^2+n}>sqrt{n},$ then
$$left|frac{1}{4sqrt{n^2+n}+2}right|<left|frac{1}{4sqrt{n}}right|<epsilonquadtext{ if }quad n>frac{1}{16epsilon^2}.$$



Would this be correct?




(*) maybe a simpler way would be to recognize that the numerator is $|s_n|<1?$






real-analysis sequences-and-series limits proof-verification







share|cite|improve this question











$endgroup$












  • $begingroup$
    Abit lost at your first expression after =, why don’t you just multiply the original by an expression to complete the difference of squares (and divide, respectively, to compensate), and then just divide numerator and denominator by $n$?
    $endgroup$
    – Makina
    Dec 15 '18 at 5:49








  • 2




    $begingroup$
    A more direct way is to note that $$sqrt{n^2+n}-n=frac1{sqrt{1+frac1n}+1}$$ whose limit is direct, as has been explained tons of times on the site (did you search for duplicates before posting this?).
    $endgroup$
    – Did
    Dec 15 '18 at 5:50










  • $begingroup$
    @Did not sure where that leads me to... I am supposed to use the definition of a limit to prove the statement, not just compute it. I am able to compute the limit, but that's not the scope here
    $endgroup$
    – DMH16
    Dec 15 '18 at 5:53












  • $begingroup$
    @Makina that's what I did, but then dividing by $n$ as mentioned by Did does not really help a whole lot, since I am not trying to compute the limit with limit properties, but show that the equality holds using the definition
    $endgroup$
    – DMH16
    Dec 15 '18 at 5:54








  • 1




    $begingroup$
    @DMH16 Your algebra and logic are correct, what part of the proof are you uncertain about? Can you justify the use of inequalities to yourself?
    $endgroup$
    – DanielV
    Dec 15 '18 at 11:13














2












2








2





$begingroup$



Prove using the definition of a limit that
$$displaystyle{lim_{ntoinfty}underbrace{sqrt{n^2+n}-n}_{s_n}}=frac{1}{2}.$$




Proof:



Let $epsilon>0$ and $ninmathbb{N}$, then
$$left|sqrt{n^2+n}-n-frac12right|stackrel{(*)}{=}left|frac{n-sqrt{n^2+n}}{2(n+sqrt{n^2+n})}right|=left|frac{n}{2(n+sqrt{n^2+n})^2}right|=left|frac{n}{4n^2+4nsqrt{n^2+n}+2n}right|<left|frac{n}{4nsqrt{n^2+n}+2n}right|=left|frac{1}{4sqrt{n^2+n}+2}right|$$
Since $forall ninmathbb{N},,sqrt{n^2+n}>sqrt{n},$ then
$$left|frac{1}{4sqrt{n^2+n}+2}right|<left|frac{1}{4sqrt{n}}right|<epsilonquadtext{ if }quad n>frac{1}{16epsilon^2}.$$



Would this be correct?




(*) maybe a simpler way would be to recognize that the numerator is $|s_n|<1?$






real-analysis sequences-and-series limits proof-verification







share|cite|improve this question











$endgroup$





Prove using the definition of a limit that
$$displaystyle{lim_{ntoinfty}underbrace{sqrt{n^2+n}-n}_{s_n}}=frac{1}{2}.$$




Proof:



Let $epsilon>0$ and $ninmathbb{N}$, then
$$left|sqrt{n^2+n}-n-frac12right|stackrel{(*)}{=}left|frac{n-sqrt{n^2+n}}{2(n+sqrt{n^2+n})}right|=left|frac{n}{2(n+sqrt{n^2+n})^2}right|=left|frac{n}{4n^2+4nsqrt{n^2+n}+2n}right|<left|frac{n}{4nsqrt{n^2+n}+2n}right|=left|frac{1}{4sqrt{n^2+n}+2}right|$$
Since $forall ninmathbb{N},,sqrt{n^2+n}>sqrt{n},$ then
$$left|frac{1}{4sqrt{n^2+n}+2}right|<left|frac{1}{4sqrt{n}}right|<epsilonquadtext{ if }quad n>frac{1}{16epsilon^2}.$$



Would this be correct?




(*) maybe a simpler way would be to recognize that the numerator is $|s_n|<1?$






real-analysis sequences-and-series limits proof-verification




real-analysis sequences-and-series limits proof-verification






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 15 '18 at 7:58







DMH16

















asked Dec 15 '18 at 5:38









DMH16DMH16

578217




578217












  • $begingroup$
    Abit lost at your first expression after =, why don’t you just multiply the original by an expression to complete the difference of squares (and divide, respectively, to compensate), and then just divide numerator and denominator by $n$?
    $endgroup$
    – Makina
    Dec 15 '18 at 5:49








  • 2




    $begingroup$
    A more direct way is to note that $$sqrt{n^2+n}-n=frac1{sqrt{1+frac1n}+1}$$ whose limit is direct, as has been explained tons of times on the site (did you search for duplicates before posting this?).
    $endgroup$
    – Did
    Dec 15 '18 at 5:50










  • $begingroup$
    @Did not sure where that leads me to... I am supposed to use the definition of a limit to prove the statement, not just compute it. I am able to compute the limit, but that's not the scope here
    $endgroup$
    – DMH16
    Dec 15 '18 at 5:53












  • $begingroup$
    @Makina that's what I did, but then dividing by $n$ as mentioned by Did does not really help a whole lot, since I am not trying to compute the limit with limit properties, but show that the equality holds using the definition
    $endgroup$
    – DMH16
    Dec 15 '18 at 5:54








  • 1




    $begingroup$
    @DMH16 Your algebra and logic are correct, what part of the proof are you uncertain about? Can you justify the use of inequalities to yourself?
    $endgroup$
    – DanielV
    Dec 15 '18 at 11:13


















  • $begingroup$
    Abit lost at your first expression after =, why don’t you just multiply the original by an expression to complete the difference of squares (and divide, respectively, to compensate), and then just divide numerator and denominator by $n$?
    $endgroup$
    – Makina
    Dec 15 '18 at 5:49








  • 2




    $begingroup$
    A more direct way is to note that $$sqrt{n^2+n}-n=frac1{sqrt{1+frac1n}+1}$$ whose limit is direct, as has been explained tons of times on the site (did you search for duplicates before posting this?).
    $endgroup$
    – Did
    Dec 15 '18 at 5:50










  • $begingroup$
    @Did not sure where that leads me to... I am supposed to use the definition of a limit to prove the statement, not just compute it. I am able to compute the limit, but that's not the scope here
    $endgroup$
    – DMH16
    Dec 15 '18 at 5:53












  • $begingroup$
    @Makina that's what I did, but then dividing by $n$ as mentioned by Did does not really help a whole lot, since I am not trying to compute the limit with limit properties, but show that the equality holds using the definition
    $endgroup$
    – DMH16
    Dec 15 '18 at 5:54








  • 1




    $begingroup$
    @DMH16 Your algebra and logic are correct, what part of the proof are you uncertain about? Can you justify the use of inequalities to yourself?
    $endgroup$
    – DanielV
    Dec 15 '18 at 11:13
















$begingroup$
Abit lost at your first expression after =, why don’t you just multiply the original by an expression to complete the difference of squares (and divide, respectively, to compensate), and then just divide numerator and denominator by $n$?
$endgroup$
– Makina
Dec 15 '18 at 5:49






$begingroup$
Abit lost at your first expression after =, why don’t you just multiply the original by an expression to complete the difference of squares (and divide, respectively, to compensate), and then just divide numerator and denominator by $n$?
$endgroup$
– Makina
Dec 15 '18 at 5:49






2




2




$begingroup$
A more direct way is to note that $$sqrt{n^2+n}-n=frac1{sqrt{1+frac1n}+1}$$ whose limit is direct, as has been explained tons of times on the site (did you search for duplicates before posting this?).
$endgroup$
– Did
Dec 15 '18 at 5:50




$begingroup$
A more direct way is to note that $$sqrt{n^2+n}-n=frac1{sqrt{1+frac1n}+1}$$ whose limit is direct, as has been explained tons of times on the site (did you search for duplicates before posting this?).
$endgroup$
– Did
Dec 15 '18 at 5:50












$begingroup$
@Did not sure where that leads me to... I am supposed to use the definition of a limit to prove the statement, not just compute it. I am able to compute the limit, but that's not the scope here
$endgroup$
– DMH16
Dec 15 '18 at 5:53






$begingroup$
@Did not sure where that leads me to... I am supposed to use the definition of a limit to prove the statement, not just compute it. I am able to compute the limit, but that's not the scope here
$endgroup$
– DMH16
Dec 15 '18 at 5:53














$begingroup$
@Makina that's what I did, but then dividing by $n$ as mentioned by Did does not really help a whole lot, since I am not trying to compute the limit with limit properties, but show that the equality holds using the definition
$endgroup$
– DMH16
Dec 15 '18 at 5:54






$begingroup$
@Makina that's what I did, but then dividing by $n$ as mentioned by Did does not really help a whole lot, since I am not trying to compute the limit with limit properties, but show that the equality holds using the definition
$endgroup$
– DMH16
Dec 15 '18 at 5:54






1




1




$begingroup$
@DMH16 Your algebra and logic are correct, what part of the proof are you uncertain about? Can you justify the use of inequalities to yourself?
$endgroup$
– DanielV
Dec 15 '18 at 11:13




$begingroup$
@DMH16 Your algebra and logic are correct, what part of the proof are you uncertain about? Can you justify the use of inequalities to yourself?
$endgroup$
– DanielV
Dec 15 '18 at 11:13










2 Answers
2






active

oldest

votes


















1












$begingroup$

We know that $$sqrt{n^2+n}-n{=(sqrt{n^2+n}-n){sqrt{n^2+n}+nover sqrt{n^2+n}+n}\={nover sqrt{n^2+n}+n}\={1over 1+sqrt{1+{1over n}}}}$$therefore$$left|{1over 1+sqrt{1+{1over n}}}-{1over 2}right|{={1over 2}-{1over 1+sqrt{1+{1over n}}}<epsilon}$$which means that $${ 1+sqrt{1+{1over n}}}<{1over {1over 2}-epsilon}\sqrt{1+{1over n}}<{{1over 2}+epsilonover {1over 2}-epsilon}\{1+{1over n}}<{{1over 4}+epsilon^2+epsilonover {1over 4}+epsilon^2-epsilon}\{1over n}<{2epsilonover {1over 4}+epsilon^2-epsilon}$$therefore by choosing $$n>{{1over 4}+epsilon^2-epsilonover 2epsilon}$$or even $n>{1over 8epsilon}$ for small enough $epsilon$ we obtain $$left|{1over 1+sqrt{1+{1over n}}}-{1over 2}right|<epsilon$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for the answer. I agree with your work, but would my procedure be correct? My question is whether my proof is correct or not.
    $endgroup$
    – DMH16
    Dec 15 '18 at 9:51






  • 1




    $begingroup$
    Yes. Quite right to me since despite of correction of your proof (though it is correct), the bound you attained to, overtakes mine :)
    $endgroup$
    – Mostafa Ayaz
    Dec 15 '18 at 11:41





















0












$begingroup$

To do it by definition, we want to prove that,



$forall varepsilon>0 exists Ninmathbb N$ such that $forall n>N$
$$Big|sqrt{n^2+n}-n-frac{1}{2}Big|<varepsilon$$
i.e.
$$Big|frac{(2n+1)-2sqrt{n^2+n}}{2}Big|<varepsilon$$
Now, by Archimedean property, $$exists Nin mathbb N : varepsilon>frac{1}{N}$$
By rationalisation, we get given expression



$=frac{1}{2(2n+1+2sqrt{n^2+n})}<frac{1}{n}<frac{1}{N}<varepsilon forall n>N$



Hope it helps






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Not sure what differs from my solution, since it appears you used the same logic, but omitted in writing certain steps. Also, I don’t think mentioning Archimedes Principle is necessary, since the definition of a limit at infinity is defined in that way.
    $endgroup$
    – DMH16
    Dec 15 '18 at 9:49










  • $begingroup$
    No, it differs significantly from yours. I am rationalising the numerator directly but you are not.
    $endgroup$
    – Martund
    Dec 15 '18 at 12:35











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

We know that $$sqrt{n^2+n}-n{=(sqrt{n^2+n}-n){sqrt{n^2+n}+nover sqrt{n^2+n}+n}\={nover sqrt{n^2+n}+n}\={1over 1+sqrt{1+{1over n}}}}$$therefore$$left|{1over 1+sqrt{1+{1over n}}}-{1over 2}right|{={1over 2}-{1over 1+sqrt{1+{1over n}}}<epsilon}$$which means that $${ 1+sqrt{1+{1over n}}}<{1over {1over 2}-epsilon}\sqrt{1+{1over n}}<{{1over 2}+epsilonover {1over 2}-epsilon}\{1+{1over n}}<{{1over 4}+epsilon^2+epsilonover {1over 4}+epsilon^2-epsilon}\{1over n}<{2epsilonover {1over 4}+epsilon^2-epsilon}$$therefore by choosing $$n>{{1over 4}+epsilon^2-epsilonover 2epsilon}$$or even $n>{1over 8epsilon}$ for small enough $epsilon$ we obtain $$left|{1over 1+sqrt{1+{1over n}}}-{1over 2}right|<epsilon$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for the answer. I agree with your work, but would my procedure be correct? My question is whether my proof is correct or not.
    $endgroup$
    – DMH16
    Dec 15 '18 at 9:51






  • 1




    $begingroup$
    Yes. Quite right to me since despite of correction of your proof (though it is correct), the bound you attained to, overtakes mine :)
    $endgroup$
    – Mostafa Ayaz
    Dec 15 '18 at 11:41


















1












$begingroup$

We know that $$sqrt{n^2+n}-n{=(sqrt{n^2+n}-n){sqrt{n^2+n}+nover sqrt{n^2+n}+n}\={nover sqrt{n^2+n}+n}\={1over 1+sqrt{1+{1over n}}}}$$therefore$$left|{1over 1+sqrt{1+{1over n}}}-{1over 2}right|{={1over 2}-{1over 1+sqrt{1+{1over n}}}<epsilon}$$which means that $${ 1+sqrt{1+{1over n}}}<{1over {1over 2}-epsilon}\sqrt{1+{1over n}}<{{1over 2}+epsilonover {1over 2}-epsilon}\{1+{1over n}}<{{1over 4}+epsilon^2+epsilonover {1over 4}+epsilon^2-epsilon}\{1over n}<{2epsilonover {1over 4}+epsilon^2-epsilon}$$therefore by choosing $$n>{{1over 4}+epsilon^2-epsilonover 2epsilon}$$or even $n>{1over 8epsilon}$ for small enough $epsilon$ we obtain $$left|{1over 1+sqrt{1+{1over n}}}-{1over 2}right|<epsilon$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for the answer. I agree with your work, but would my procedure be correct? My question is whether my proof is correct or not.
    $endgroup$
    – DMH16
    Dec 15 '18 at 9:51






  • 1




    $begingroup$
    Yes. Quite right to me since despite of correction of your proof (though it is correct), the bound you attained to, overtakes mine :)
    $endgroup$
    – Mostafa Ayaz
    Dec 15 '18 at 11:41
















1












1








1





$begingroup$

We know that $$sqrt{n^2+n}-n{=(sqrt{n^2+n}-n){sqrt{n^2+n}+nover sqrt{n^2+n}+n}\={nover sqrt{n^2+n}+n}\={1over 1+sqrt{1+{1over n}}}}$$therefore$$left|{1over 1+sqrt{1+{1over n}}}-{1over 2}right|{={1over 2}-{1over 1+sqrt{1+{1over n}}}<epsilon}$$which means that $${ 1+sqrt{1+{1over n}}}<{1over {1over 2}-epsilon}\sqrt{1+{1over n}}<{{1over 2}+epsilonover {1over 2}-epsilon}\{1+{1over n}}<{{1over 4}+epsilon^2+epsilonover {1over 4}+epsilon^2-epsilon}\{1over n}<{2epsilonover {1over 4}+epsilon^2-epsilon}$$therefore by choosing $$n>{{1over 4}+epsilon^2-epsilonover 2epsilon}$$or even $n>{1over 8epsilon}$ for small enough $epsilon$ we obtain $$left|{1over 1+sqrt{1+{1over n}}}-{1over 2}right|<epsilon$$






share|cite|improve this answer











$endgroup$



We know that $$sqrt{n^2+n}-n{=(sqrt{n^2+n}-n){sqrt{n^2+n}+nover sqrt{n^2+n}+n}\={nover sqrt{n^2+n}+n}\={1over 1+sqrt{1+{1over n}}}}$$therefore$$left|{1over 1+sqrt{1+{1over n}}}-{1over 2}right|{={1over 2}-{1over 1+sqrt{1+{1over n}}}<epsilon}$$which means that $${ 1+sqrt{1+{1over n}}}<{1over {1over 2}-epsilon}\sqrt{1+{1over n}}<{{1over 2}+epsilonover {1over 2}-epsilon}\{1+{1over n}}<{{1over 4}+epsilon^2+epsilonover {1over 4}+epsilon^2-epsilon}\{1over n}<{2epsilonover {1over 4}+epsilon^2-epsilon}$$therefore by choosing $$n>{{1over 4}+epsilon^2-epsilonover 2epsilon}$$or even $n>{1over 8epsilon}$ for small enough $epsilon$ we obtain $$left|{1over 1+sqrt{1+{1over n}}}-{1over 2}right|<epsilon$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 15 '18 at 9:19

























answered Dec 15 '18 at 9:09









Mostafa AyazMostafa Ayaz

15.2k3939




15.2k3939












  • $begingroup$
    Thanks for the answer. I agree with your work, but would my procedure be correct? My question is whether my proof is correct or not.
    $endgroup$
    – DMH16
    Dec 15 '18 at 9:51






  • 1




    $begingroup$
    Yes. Quite right to me since despite of correction of your proof (though it is correct), the bound you attained to, overtakes mine :)
    $endgroup$
    – Mostafa Ayaz
    Dec 15 '18 at 11:41




















  • $begingroup$
    Thanks for the answer. I agree with your work, but would my procedure be correct? My question is whether my proof is correct or not.
    $endgroup$
    – DMH16
    Dec 15 '18 at 9:51






  • 1




    $begingroup$
    Yes. Quite right to me since despite of correction of your proof (though it is correct), the bound you attained to, overtakes mine :)
    $endgroup$
    – Mostafa Ayaz
    Dec 15 '18 at 11:41


















$begingroup$
Thanks for the answer. I agree with your work, but would my procedure be correct? My question is whether my proof is correct or not.
$endgroup$
– DMH16
Dec 15 '18 at 9:51




$begingroup$
Thanks for the answer. I agree with your work, but would my procedure be correct? My question is whether my proof is correct or not.
$endgroup$
– DMH16
Dec 15 '18 at 9:51




1




1




$begingroup$
Yes. Quite right to me since despite of correction of your proof (though it is correct), the bound you attained to, overtakes mine :)
$endgroup$
– Mostafa Ayaz
Dec 15 '18 at 11:41






$begingroup$
Yes. Quite right to me since despite of correction of your proof (though it is correct), the bound you attained to, overtakes mine :)
$endgroup$
– Mostafa Ayaz
Dec 15 '18 at 11:41













0












$begingroup$

To do it by definition, we want to prove that,



$forall varepsilon>0 exists Ninmathbb N$ such that $forall n>N$
$$Big|sqrt{n^2+n}-n-frac{1}{2}Big|<varepsilon$$
i.e.
$$Big|frac{(2n+1)-2sqrt{n^2+n}}{2}Big|<varepsilon$$
Now, by Archimedean property, $$exists Nin mathbb N : varepsilon>frac{1}{N}$$
By rationalisation, we get given expression



$=frac{1}{2(2n+1+2sqrt{n^2+n})}<frac{1}{n}<frac{1}{N}<varepsilon forall n>N$



Hope it helps






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Not sure what differs from my solution, since it appears you used the same logic, but omitted in writing certain steps. Also, I don’t think mentioning Archimedes Principle is necessary, since the definition of a limit at infinity is defined in that way.
    $endgroup$
    – DMH16
    Dec 15 '18 at 9:49










  • $begingroup$
    No, it differs significantly from yours. I am rationalising the numerator directly but you are not.
    $endgroup$
    – Martund
    Dec 15 '18 at 12:35
















0












$begingroup$

To do it by definition, we want to prove that,



$forall varepsilon>0 exists Ninmathbb N$ such that $forall n>N$
$$Big|sqrt{n^2+n}-n-frac{1}{2}Big|<varepsilon$$
i.e.
$$Big|frac{(2n+1)-2sqrt{n^2+n}}{2}Big|<varepsilon$$
Now, by Archimedean property, $$exists Nin mathbb N : varepsilon>frac{1}{N}$$
By rationalisation, we get given expression



$=frac{1}{2(2n+1+2sqrt{n^2+n})}<frac{1}{n}<frac{1}{N}<varepsilon forall n>N$



Hope it helps






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Not sure what differs from my solution, since it appears you used the same logic, but omitted in writing certain steps. Also, I don’t think mentioning Archimedes Principle is necessary, since the definition of a limit at infinity is defined in that way.
    $endgroup$
    – DMH16
    Dec 15 '18 at 9:49










  • $begingroup$
    No, it differs significantly from yours. I am rationalising the numerator directly but you are not.
    $endgroup$
    – Martund
    Dec 15 '18 at 12:35














0












0








0





$begingroup$

To do it by definition, we want to prove that,



$forall varepsilon>0 exists Ninmathbb N$ such that $forall n>N$
$$Big|sqrt{n^2+n}-n-frac{1}{2}Big|<varepsilon$$
i.e.
$$Big|frac{(2n+1)-2sqrt{n^2+n}}{2}Big|<varepsilon$$
Now, by Archimedean property, $$exists Nin mathbb N : varepsilon>frac{1}{N}$$
By rationalisation, we get given expression



$=frac{1}{2(2n+1+2sqrt{n^2+n})}<frac{1}{n}<frac{1}{N}<varepsilon forall n>N$



Hope it helps






share|cite|improve this answer











$endgroup$



To do it by definition, we want to prove that,



$forall varepsilon>0 exists Ninmathbb N$ such that $forall n>N$
$$Big|sqrt{n^2+n}-n-frac{1}{2}Big|<varepsilon$$
i.e.
$$Big|frac{(2n+1)-2sqrt{n^2+n}}{2}Big|<varepsilon$$
Now, by Archimedean property, $$exists Nin mathbb N : varepsilon>frac{1}{N}$$
By rationalisation, we get given expression



$=frac{1}{2(2n+1+2sqrt{n^2+n})}<frac{1}{n}<frac{1}{N}<varepsilon forall n>N$



Hope it helps







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 15 '18 at 10:16

























answered Dec 15 '18 at 9:34









MartundMartund

1,413212




1,413212












  • $begingroup$
    Not sure what differs from my solution, since it appears you used the same logic, but omitted in writing certain steps. Also, I don’t think mentioning Archimedes Principle is necessary, since the definition of a limit at infinity is defined in that way.
    $endgroup$
    – DMH16
    Dec 15 '18 at 9:49










  • $begingroup$
    No, it differs significantly from yours. I am rationalising the numerator directly but you are not.
    $endgroup$
    – Martund
    Dec 15 '18 at 12:35


















  • $begingroup$
    Not sure what differs from my solution, since it appears you used the same logic, but omitted in writing certain steps. Also, I don’t think mentioning Archimedes Principle is necessary, since the definition of a limit at infinity is defined in that way.
    $endgroup$
    – DMH16
    Dec 15 '18 at 9:49










  • $begingroup$
    No, it differs significantly from yours. I am rationalising the numerator directly but you are not.
    $endgroup$
    – Martund
    Dec 15 '18 at 12:35
















$begingroup$
Not sure what differs from my solution, since it appears you used the same logic, but omitted in writing certain steps. Also, I don’t think mentioning Archimedes Principle is necessary, since the definition of a limit at infinity is defined in that way.
$endgroup$
– DMH16
Dec 15 '18 at 9:49




$begingroup$
Not sure what differs from my solution, since it appears you used the same logic, but omitted in writing certain steps. Also, I don’t think mentioning Archimedes Principle is necessary, since the definition of a limit at infinity is defined in that way.
$endgroup$
– DMH16
Dec 15 '18 at 9:49












$begingroup$
No, it differs significantly from yours. I am rationalising the numerator directly but you are not.
$endgroup$
– Martund
Dec 15 '18 at 12:35




$begingroup$
No, it differs significantly from yours. I am rationalising the numerator directly but you are not.
$endgroup$
– Martund
Dec 15 '18 at 12:35


















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