Limits: Why can you sometimes factor out the highest exponent under a root, and then let the root evaluate to...












0














I'm supposed to solve this exercise:
1



Now I also have the (logically seeming) laws from the lecture slides:
2 ° 3



So at first i though you could just rewrite the term by factoring out the n^4 below the root and and then let everything inside the root except the 1 evaluate to 0 by taking the limit (and then let the n^2 cancel each other out and get 1 as result).
4



This is wrong (altough i'm not sure yet in which way/how the formulas were misapplicated). I also have the solution to the exercise:
5



My central understanding problem is in the last line of the solution: Why is it correct to let take individual limits inside the root and let everything (except the 1) evaluate to 0 in this situation, but not in the "original" formula (as i did)?



Thanks for any help




original title (if that’s more understable) (without character-limit):
Limits: Why can you sometimes factor out the highest exponent under a root, and then let everything else inside the root evaluate to 0 when taking the limit, but othertimes (apparently) not?






limits analysis







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    0














    I'm supposed to solve this exercise:
    1



    Now I also have the (logically seeming) laws from the lecture slides:
    2 ° 3



    So at first i though you could just rewrite the term by factoring out the n^4 below the root and and then let everything inside the root except the 1 evaluate to 0 by taking the limit (and then let the n^2 cancel each other out and get 1 as result).
    4



    This is wrong (altough i'm not sure yet in which way/how the formulas were misapplicated). I also have the solution to the exercise:
    5



    My central understanding problem is in the last line of the solution: Why is it correct to let take individual limits inside the root and let everything (except the 1) evaluate to 0 in this situation, but not in the "original" formula (as i did)?



    Thanks for any help




    original title (if that’s more understable) (without character-limit):
    Limits: Why can you sometimes factor out the highest exponent under a root, and then let everything else inside the root evaluate to 0 when taking the limit, but othertimes (apparently) not?






    limits analysis







    share|cite|improve this question







    New contributor




    user631488 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.























      0












      0








      0







      I'm supposed to solve this exercise:
      1



      Now I also have the (logically seeming) laws from the lecture slides:
      2 ° 3



      So at first i though you could just rewrite the term by factoring out the n^4 below the root and and then let everything inside the root except the 1 evaluate to 0 by taking the limit (and then let the n^2 cancel each other out and get 1 as result).
      4



      This is wrong (altough i'm not sure yet in which way/how the formulas were misapplicated). I also have the solution to the exercise:
      5



      My central understanding problem is in the last line of the solution: Why is it correct to let take individual limits inside the root and let everything (except the 1) evaluate to 0 in this situation, but not in the "original" formula (as i did)?



      Thanks for any help




      original title (if that’s more understable) (without character-limit):
      Limits: Why can you sometimes factor out the highest exponent under a root, and then let everything else inside the root evaluate to 0 when taking the limit, but othertimes (apparently) not?






      limits analysis







      share|cite|improve this question







      New contributor




      user631488 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      I'm supposed to solve this exercise:
      1



      Now I also have the (logically seeming) laws from the lecture slides:
      2 ° 3



      So at first i though you could just rewrite the term by factoring out the n^4 below the root and and then let everything inside the root except the 1 evaluate to 0 by taking the limit (and then let the n^2 cancel each other out and get 1 as result).
      4



      This is wrong (altough i'm not sure yet in which way/how the formulas were misapplicated). I also have the solution to the exercise:
      5



      My central understanding problem is in the last line of the solution: Why is it correct to let take individual limits inside the root and let everything (except the 1) evaluate to 0 in this situation, but not in the "original" formula (as i did)?



      Thanks for any help




      original title (if that’s more understable) (without character-limit):
      Limits: Why can you sometimes factor out the highest exponent under a root, and then let everything else inside the root evaluate to 0 when taking the limit, but othertimes (apparently) not?






      limits analysis




      limits analysis






      share|cite|improve this question







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      user631488 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite|improve this question







      New contributor




      user631488 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









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      asked Jan 4 at 0:27









      user631488

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          1 Answer
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          active

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          2














          The rules (2 and 3) you cited have specific hypotheses, which can be easy to overlook when remembering formulae. We know that, if $a_n to a$ and $b_n to b$ ($a$ and $b$ finite) as $n to infty$, then
          $$a_n pm b_n to a pm b.$$
          Similar rules apply for multiplication, and even division if care is taken to make sure no division by $0$ happens on either side. There's also the rule for powers, as you've stated. If $a_n to a$, with $a_n, a ge 0$ and $r in mathbb{R}$, then
          $$a_n^r to a^r.$$
          Note that all of these rules require you to have convergent sequences to start off with.



          In your working, you write




          $$lim (n^2 cdot sqrt{1 + n^{-2} + n^{-3} + n^{-4}} - n^2 + 1) = lim (n^2 cdot 1 - n^2 + 1).$$




          This step, replacing $sqrt{1 + n^{-2} + n^{-3} + n^{-4}}$ by its limit, is not supported by the above rules. If $n^2$ were also convergent (to finite $L$, say), then you could simultaneously replace it by its limit $L$, to produce the working,



          $$lim (n^2 cdot sqrt{1 + n^{-2} + n^{-3} + n^{-4}} - n^2 + 1) = L cdot 1 - L + 1 = 1.$$



          This would be allowed, as each sequence is convergent to some finite limit, and the rules apply. But, $n^2$ does not approach any finite limit! The limit laws do not apply. You have to use a method like the solutions, or apply L'Hopital's rule, or something else a little more clever.






          share|cite|improve this answer





















          • Thanks for the answer. (Really missed a central aspect (of the topic), so thanks a lot for the detailed explaination and for making it general (as well as relating it to the original question) (Similar rules apply for multiplication, and even division if care is taken to make sure no division by 0 happens on either side. Note that all of these rules require you to have convergent sequences to start off with. etc)).
            – user631488
            Jan 4 at 1:18











          Your Answer





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          1 Answer
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          active

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          1 Answer
          1






          active

          oldest

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          active

          oldest

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          active

          oldest

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          2














          The rules (2 and 3) you cited have specific hypotheses, which can be easy to overlook when remembering formulae. We know that, if $a_n to a$ and $b_n to b$ ($a$ and $b$ finite) as $n to infty$, then
          $$a_n pm b_n to a pm b.$$
          Similar rules apply for multiplication, and even division if care is taken to make sure no division by $0$ happens on either side. There's also the rule for powers, as you've stated. If $a_n to a$, with $a_n, a ge 0$ and $r in mathbb{R}$, then
          $$a_n^r to a^r.$$
          Note that all of these rules require you to have convergent sequences to start off with.



          In your working, you write




          $$lim (n^2 cdot sqrt{1 + n^{-2} + n^{-3} + n^{-4}} - n^2 + 1) = lim (n^2 cdot 1 - n^2 + 1).$$




          This step, replacing $sqrt{1 + n^{-2} + n^{-3} + n^{-4}}$ by its limit, is not supported by the above rules. If $n^2$ were also convergent (to finite $L$, say), then you could simultaneously replace it by its limit $L$, to produce the working,



          $$lim (n^2 cdot sqrt{1 + n^{-2} + n^{-3} + n^{-4}} - n^2 + 1) = L cdot 1 - L + 1 = 1.$$



          This would be allowed, as each sequence is convergent to some finite limit, and the rules apply. But, $n^2$ does not approach any finite limit! The limit laws do not apply. You have to use a method like the solutions, or apply L'Hopital's rule, or something else a little more clever.






          share|cite|improve this answer





















          • Thanks for the answer. (Really missed a central aspect (of the topic), so thanks a lot for the detailed explaination and for making it general (as well as relating it to the original question) (Similar rules apply for multiplication, and even division if care is taken to make sure no division by 0 happens on either side. Note that all of these rules require you to have convergent sequences to start off with. etc)).
            – user631488
            Jan 4 at 1:18
















          2














          The rules (2 and 3) you cited have specific hypotheses, which can be easy to overlook when remembering formulae. We know that, if $a_n to a$ and $b_n to b$ ($a$ and $b$ finite) as $n to infty$, then
          $$a_n pm b_n to a pm b.$$
          Similar rules apply for multiplication, and even division if care is taken to make sure no division by $0$ happens on either side. There's also the rule for powers, as you've stated. If $a_n to a$, with $a_n, a ge 0$ and $r in mathbb{R}$, then
          $$a_n^r to a^r.$$
          Note that all of these rules require you to have convergent sequences to start off with.



          In your working, you write




          $$lim (n^2 cdot sqrt{1 + n^{-2} + n^{-3} + n^{-4}} - n^2 + 1) = lim (n^2 cdot 1 - n^2 + 1).$$




          This step, replacing $sqrt{1 + n^{-2} + n^{-3} + n^{-4}}$ by its limit, is not supported by the above rules. If $n^2$ were also convergent (to finite $L$, say), then you could simultaneously replace it by its limit $L$, to produce the working,



          $$lim (n^2 cdot sqrt{1 + n^{-2} + n^{-3} + n^{-4}} - n^2 + 1) = L cdot 1 - L + 1 = 1.$$



          This would be allowed, as each sequence is convergent to some finite limit, and the rules apply. But, $n^2$ does not approach any finite limit! The limit laws do not apply. You have to use a method like the solutions, or apply L'Hopital's rule, or something else a little more clever.






          share|cite|improve this answer





















          • Thanks for the answer. (Really missed a central aspect (of the topic), so thanks a lot for the detailed explaination and for making it general (as well as relating it to the original question) (Similar rules apply for multiplication, and even division if care is taken to make sure no division by 0 happens on either side. Note that all of these rules require you to have convergent sequences to start off with. etc)).
            – user631488
            Jan 4 at 1:18














          2












          2








          2






          The rules (2 and 3) you cited have specific hypotheses, which can be easy to overlook when remembering formulae. We know that, if $a_n to a$ and $b_n to b$ ($a$ and $b$ finite) as $n to infty$, then
          $$a_n pm b_n to a pm b.$$
          Similar rules apply for multiplication, and even division if care is taken to make sure no division by $0$ happens on either side. There's also the rule for powers, as you've stated. If $a_n to a$, with $a_n, a ge 0$ and $r in mathbb{R}$, then
          $$a_n^r to a^r.$$
          Note that all of these rules require you to have convergent sequences to start off with.



          In your working, you write




          $$lim (n^2 cdot sqrt{1 + n^{-2} + n^{-3} + n^{-4}} - n^2 + 1) = lim (n^2 cdot 1 - n^2 + 1).$$




          This step, replacing $sqrt{1 + n^{-2} + n^{-3} + n^{-4}}$ by its limit, is not supported by the above rules. If $n^2$ were also convergent (to finite $L$, say), then you could simultaneously replace it by its limit $L$, to produce the working,



          $$lim (n^2 cdot sqrt{1 + n^{-2} + n^{-3} + n^{-4}} - n^2 + 1) = L cdot 1 - L + 1 = 1.$$



          This would be allowed, as each sequence is convergent to some finite limit, and the rules apply. But, $n^2$ does not approach any finite limit! The limit laws do not apply. You have to use a method like the solutions, or apply L'Hopital's rule, or something else a little more clever.






          share|cite|improve this answer












          The rules (2 and 3) you cited have specific hypotheses, which can be easy to overlook when remembering formulae. We know that, if $a_n to a$ and $b_n to b$ ($a$ and $b$ finite) as $n to infty$, then
          $$a_n pm b_n to a pm b.$$
          Similar rules apply for multiplication, and even division if care is taken to make sure no division by $0$ happens on either side. There's also the rule for powers, as you've stated. If $a_n to a$, with $a_n, a ge 0$ and $r in mathbb{R}$, then
          $$a_n^r to a^r.$$
          Note that all of these rules require you to have convergent sequences to start off with.



          In your working, you write




          $$lim (n^2 cdot sqrt{1 + n^{-2} + n^{-3} + n^{-4}} - n^2 + 1) = lim (n^2 cdot 1 - n^2 + 1).$$




          This step, replacing $sqrt{1 + n^{-2} + n^{-3} + n^{-4}}$ by its limit, is not supported by the above rules. If $n^2$ were also convergent (to finite $L$, say), then you could simultaneously replace it by its limit $L$, to produce the working,



          $$lim (n^2 cdot sqrt{1 + n^{-2} + n^{-3} + n^{-4}} - n^2 + 1) = L cdot 1 - L + 1 = 1.$$



          This would be allowed, as each sequence is convergent to some finite limit, and the rules apply. But, $n^2$ does not approach any finite limit! The limit laws do not apply. You have to use a method like the solutions, or apply L'Hopital's rule, or something else a little more clever.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 4 at 0:48









          Theo Bendit

          16.7k12148




          16.7k12148












          • Thanks for the answer. (Really missed a central aspect (of the topic), so thanks a lot for the detailed explaination and for making it general (as well as relating it to the original question) (Similar rules apply for multiplication, and even division if care is taken to make sure no division by 0 happens on either side. Note that all of these rules require you to have convergent sequences to start off with. etc)).
            – user631488
            Jan 4 at 1:18


















          • Thanks for the answer. (Really missed a central aspect (of the topic), so thanks a lot for the detailed explaination and for making it general (as well as relating it to the original question) (Similar rules apply for multiplication, and even division if care is taken to make sure no division by 0 happens on either side. Note that all of these rules require you to have convergent sequences to start off with. etc)).
            – user631488
            Jan 4 at 1:18
















          Thanks for the answer. (Really missed a central aspect (of the topic), so thanks a lot for the detailed explaination and for making it general (as well as relating it to the original question) (Similar rules apply for multiplication, and even division if care is taken to make sure no division by 0 happens on either side. Note that all of these rules require you to have convergent sequences to start off with. etc)).
          – user631488
          Jan 4 at 1:18




          Thanks for the answer. (Really missed a central aspect (of the topic), so thanks a lot for the detailed explaination and for making it general (as well as relating it to the original question) (Similar rules apply for multiplication, and even division if care is taken to make sure no division by 0 happens on either side. Note that all of these rules require you to have convergent sequences to start off with. etc)).
          – user631488
          Jan 4 at 1:18










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