Equality or not in definition of limits and infinity and sequences?

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











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Sequences



Limits



See in the first case, there is an equality also, between $n$ And $N$, but in the second case, there is a strict inequality between $x$ and $N$.



So what I want to ask is that if we remove the equality in the first case and add an equality in the second case, do the definitions still hold the same meaning?



And to prove the equivalence of both the definitions we would have to prove "if and only if" condition, that is, first definition implies the modified one, vice versa. So that's what I am looking for.




real-analysis sequences-and-series definition




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    Yes, those are equivalent
    – Rumpelstiltskin
    Aug 6 at 17:06














up vote
0
down vote

favorite












Sequences



Limits



See in the first case, there is an equality also, between $n$ And $N$, but in the second case, there is a strict inequality between $x$ and $N$.



So what I want to ask is that if we remove the equality in the first case and add an equality in the second case, do the definitions still hold the same meaning?



And to prove the equivalence of both the definitions we would have to prove "if and only if" condition, that is, first definition implies the modified one, vice versa. So that's what I am looking for.




real-analysis sequences-and-series definition




share|cite|improve this question

















  • 1




    Yes, those are equivalent
    – Rumpelstiltskin
    Aug 6 at 17:06












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Sequences



Limits



See in the first case, there is an equality also, between $n$ And $N$, but in the second case, there is a strict inequality between $x$ and $N$.



So what I want to ask is that if we remove the equality in the first case and add an equality in the second case, do the definitions still hold the same meaning?



And to prove the equivalence of both the definitions we would have to prove "if and only if" condition, that is, first definition implies the modified one, vice versa. So that's what I am looking for.




real-analysis sequences-and-series definition




share|cite|improve this question













Sequences



Limits



See in the first case, there is an equality also, between $n$ And $N$, but in the second case, there is a strict inequality between $x$ and $N$.



So what I want to ask is that if we remove the equality in the first case and add an equality in the second case, do the definitions still hold the same meaning?



And to prove the equivalence of both the definitions we would have to prove "if and only if" condition, that is, first definition implies the modified one, vice versa. So that's what I am looking for.





real-analysis sequences-and-series definition





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Aug 6 at 17:06
























asked Aug 6 at 16:57









Aditya Agarwal

2,93111536




2,93111536







  • 1




    Yes, those are equivalent
    – Rumpelstiltskin
    Aug 6 at 17:06












  • 1




    Yes, those are equivalent
    – Rumpelstiltskin
    Aug 6 at 17:06







1




1




Yes, those are equivalent
– Rumpelstiltskin
Aug 6 at 17:06




Yes, those are equivalent
– Rumpelstiltskin
Aug 6 at 17:06










1 Answer
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For fixed $epsilon>0$, if such an $N(epsilon)$ exists in the functional definition, then define $N':=N+1$, so that the result holds for $xgeq N'$. In other words, equality is immaterial to the definition.






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  • Got ye! Thanks.
    – Aditya Agarwal
    Aug 6 at 17:15










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1 Answer
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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
1
down vote



accepted










For fixed $epsilon>0$, if such an $N(epsilon)$ exists in the functional definition, then define $N':=N+1$, so that the result holds for $xgeq N'$. In other words, equality is immaterial to the definition.






share|cite|improve this answer





















  • Got ye! Thanks.
    – Aditya Agarwal
    Aug 6 at 17:15














up vote
1
down vote



accepted










For fixed $epsilon>0$, if such an $N(epsilon)$ exists in the functional definition, then define $N':=N+1$, so that the result holds for $xgeq N'$. In other words, equality is immaterial to the definition.






share|cite|improve this answer





















  • Got ye! Thanks.
    – Aditya Agarwal
    Aug 6 at 17:15












up vote
1
down vote



accepted







up vote
1
down vote



accepted






For fixed $epsilon>0$, if such an $N(epsilon)$ exists in the functional definition, then define $N':=N+1$, so that the result holds for $xgeq N'$. In other words, equality is immaterial to the definition.






share|cite|improve this answer













For fixed $epsilon>0$, if such an $N(epsilon)$ exists in the functional definition, then define $N':=N+1$, so that the result holds for $xgeq N'$. In other words, equality is immaterial to the definition.







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answered Aug 6 at 17:06









Alex R.

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  • Got ye! Thanks.
    – Aditya Agarwal
    Aug 6 at 17:15
















  • Got ye! Thanks.
    – Aditya Agarwal
    Aug 6 at 17:15















Got ye! Thanks.
– Aditya Agarwal
Aug 6 at 17:15




Got ye! Thanks.
– Aditya Agarwal
Aug 6 at 17:15












 

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