Proving of limits

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  1. Let $a>0$ and $n$ be a positive integer. Prove that $$lim_xto a x^1/n = a^1/n$$

For this question, I'm not entirely sure whether I should proceed with the $epsilon$-$delta$ definition of a limit, or simply prove the power law for limits, since the question did not specify the need to use the precise definition of a limit in this proof. I'm thinking that I could try to prove that the function is continuous at $x=a$, but proving continuity requires $lim_xto a x^1/n = a^1/n$. Could someone hint me in the right direction for this?




  1. Let $a$ be a real number. Prove that



    (a) $lim_xto a^- frac1(x-a)^3 = -infty$ and



    (b)$lim_xto a^+ frac1(x-a)^3 = infty$



My attempt at the second question is as follows:



(a) We want to prove $forall N<0$, $exists delta >0$ such that if $a - delta < x < a$, then $frac1(x-a)^3 < N$. $$frac1(x-a)^3<N$$ $$frac1N > (x-a)^3$$ so $x-a > sqrt[3]frac1N$, i.e. $x > sqrt[3]frac1N$, so we choose $delta = -sqrt[3]frac1N$. Note that $sqrt[3]frac1N + a < x < a$, so $frac1(x-a)^3 < N$ (Q.E.D)



(b) Similarly, we want to prove that $forall M > 0$, $exists delta >0$ such that if $a<x<a+delta$ then $frac1(x-a)^3 > M$. $$frac1M > (x-a)^3$$ $$x < sqrt[3]frac1M + a$$, so we choose $delta = sqrt[3]frac1M$. Note that $a<x<a+ sqrt[3]frac1M$, so $frac1(x-a)^3 > M$ (Q.E.D)



I feel like this proof I produced is missing something necessary to conclude the respective limits, or it could be completely wrong, so any criticism for the proof would be appreciated!



Thanks in advance!










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    up vote
    2
    down vote

    favorite












    1. Let $a>0$ and $n$ be a positive integer. Prove that $$lim_xto a x^1/n = a^1/n$$

    For this question, I'm not entirely sure whether I should proceed with the $epsilon$-$delta$ definition of a limit, or simply prove the power law for limits, since the question did not specify the need to use the precise definition of a limit in this proof. I'm thinking that I could try to prove that the function is continuous at $x=a$, but proving continuity requires $lim_xto a x^1/n = a^1/n$. Could someone hint me in the right direction for this?




    1. Let $a$ be a real number. Prove that



      (a) $lim_xto a^- frac1(x-a)^3 = -infty$ and



      (b)$lim_xto a^+ frac1(x-a)^3 = infty$



    My attempt at the second question is as follows:



    (a) We want to prove $forall N<0$, $exists delta >0$ such that if $a - delta < x < a$, then $frac1(x-a)^3 < N$. $$frac1(x-a)^3<N$$ $$frac1N > (x-a)^3$$ so $x-a > sqrt[3]frac1N$, i.e. $x > sqrt[3]frac1N$, so we choose $delta = -sqrt[3]frac1N$. Note that $sqrt[3]frac1N + a < x < a$, so $frac1(x-a)^3 < N$ (Q.E.D)



    (b) Similarly, we want to prove that $forall M > 0$, $exists delta >0$ such that if $a<x<a+delta$ then $frac1(x-a)^3 > M$. $$frac1M > (x-a)^3$$ $$x < sqrt[3]frac1M + a$$, so we choose $delta = sqrt[3]frac1M$. Note that $a<x<a+ sqrt[3]frac1M$, so $frac1(x-a)^3 > M$ (Q.E.D)



    I feel like this proof I produced is missing something necessary to conclude the respective limits, or it could be completely wrong, so any criticism for the proof would be appreciated!



    Thanks in advance!










    share|cite|improve this question

























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      1. Let $a>0$ and $n$ be a positive integer. Prove that $$lim_xto a x^1/n = a^1/n$$

      For this question, I'm not entirely sure whether I should proceed with the $epsilon$-$delta$ definition of a limit, or simply prove the power law for limits, since the question did not specify the need to use the precise definition of a limit in this proof. I'm thinking that I could try to prove that the function is continuous at $x=a$, but proving continuity requires $lim_xto a x^1/n = a^1/n$. Could someone hint me in the right direction for this?




      1. Let $a$ be a real number. Prove that



        (a) $lim_xto a^- frac1(x-a)^3 = -infty$ and



        (b)$lim_xto a^+ frac1(x-a)^3 = infty$



      My attempt at the second question is as follows:



      (a) We want to prove $forall N<0$, $exists delta >0$ such that if $a - delta < x < a$, then $frac1(x-a)^3 < N$. $$frac1(x-a)^3<N$$ $$frac1N > (x-a)^3$$ so $x-a > sqrt[3]frac1N$, i.e. $x > sqrt[3]frac1N$, so we choose $delta = -sqrt[3]frac1N$. Note that $sqrt[3]frac1N + a < x < a$, so $frac1(x-a)^3 < N$ (Q.E.D)



      (b) Similarly, we want to prove that $forall M > 0$, $exists delta >0$ such that if $a<x<a+delta$ then $frac1(x-a)^3 > M$. $$frac1M > (x-a)^3$$ $$x < sqrt[3]frac1M + a$$, so we choose $delta = sqrt[3]frac1M$. Note that $a<x<a+ sqrt[3]frac1M$, so $frac1(x-a)^3 > M$ (Q.E.D)



      I feel like this proof I produced is missing something necessary to conclude the respective limits, or it could be completely wrong, so any criticism for the proof would be appreciated!



      Thanks in advance!










      share|cite|improve this question















      1. Let $a>0$ and $n$ be a positive integer. Prove that $$lim_xto a x^1/n = a^1/n$$

      For this question, I'm not entirely sure whether I should proceed with the $epsilon$-$delta$ definition of a limit, or simply prove the power law for limits, since the question did not specify the need to use the precise definition of a limit in this proof. I'm thinking that I could try to prove that the function is continuous at $x=a$, but proving continuity requires $lim_xto a x^1/n = a^1/n$. Could someone hint me in the right direction for this?




      1. Let $a$ be a real number. Prove that



        (a) $lim_xto a^- frac1(x-a)^3 = -infty$ and



        (b)$lim_xto a^+ frac1(x-a)^3 = infty$



      My attempt at the second question is as follows:



      (a) We want to prove $forall N<0$, $exists delta >0$ such that if $a - delta < x < a$, then $frac1(x-a)^3 < N$. $$frac1(x-a)^3<N$$ $$frac1N > (x-a)^3$$ so $x-a > sqrt[3]frac1N$, i.e. $x > sqrt[3]frac1N$, so we choose $delta = -sqrt[3]frac1N$. Note that $sqrt[3]frac1N + a < x < a$, so $frac1(x-a)^3 < N$ (Q.E.D)



      (b) Similarly, we want to prove that $forall M > 0$, $exists delta >0$ such that if $a<x<a+delta$ then $frac1(x-a)^3 > M$. $$frac1M > (x-a)^3$$ $$x < sqrt[3]frac1M + a$$, so we choose $delta = sqrt[3]frac1M$. Note that $a<x<a+ sqrt[3]frac1M$, so $frac1(x-a)^3 > M$ (Q.E.D)



      I feel like this proof I produced is missing something necessary to conclude the respective limits, or it could be completely wrong, so any criticism for the proof would be appreciated!



      Thanks in advance!







      calculus limits






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      edited Sep 9 at 11:07

























      asked Sep 9 at 11:00









      uznam

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          2 Answers
          2






          active

          oldest

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          up vote
          1
          down vote



          accepted










          For question 1, use the following geometric series identity:
          $$fracp^n - q^np - q = p^n-1 + p^n - 2 q + p^n - 3 q^2 + ldots + q^n - 1$$
          with $p = x^frac1n$ and $q = a^frac1n$.



          As for question 2, you're doing fine. There is what I think is an inconsequential typo under your first two inequalities, where you say $x > sqrt[3]frac1N$, where I think you may mean $x > a + sqrt[3]frac1N$, but it doesn't cause any problems down the road.



          It's hard for me to gauge exactly what is bothering you; I can only guess. However, what slightly bothers me is that the proof is presented slightly backwards. It's presented in the order of your working out, rather than in logical order.



          You're supposed to be able to take a number $N < 0$, and instantly produce a $delta > 0$ with the given property. So, your proof should probably take the form:




          Suppose $N < 0$, and let $delta = -sqrt[3]frac1N > 0$. Then $a - delta < x < a implies ldots implies frac1(x - a)^3 < N.$




          The working you've written is valuable working behind the scenes, but it's nicer to present the proof in the proper order.



          EDIT: In response to your comment, that's not too hard to do, and I think it is worth covering. Anything you feel is shaky is worth delving into. A proof is supposed to be convincing, and if it can't even convince the person who wrote it, then it's not working.



          To prove this, suppose that $a + sqrt[3]frac1N < x < a$. Then
          $$x - a > sqrt[3]frac1N.$$
          Then, because the function $x mapsto x^3$ is increasing, we have
          $$(x - a)^3 > left(sqrt[3]frac1Nright)^3 = frac1N.$$
          Now, $x < a$, so $x - a < 0$. For negative numbers, the map $x mapsto frac1x$ is decreasing, so
          $$frac1(x - a)^3 < N,$$
          which is what we wanted to prove.






          share|cite|improve this answer






















          • Thank you for your response. To clarify what exactly was bothering me about my proof: I feel like there could be some way to explain how we conclude that $frac1(x-a)^3 < N$ from the premise that $sqrt[3]frac1N + a < x < a$ for part (a), and likewise for part (b). Or is it unnecessary to explain this since it is just a simple fact from mere observation?
            – uznam
            Sep 9 at 13:36







          • 1




            @uznam Edited. Hope it helps.
            – Theo Bendit
            Sep 9 at 14:37

















          up vote
          1
          down vote













          For $(1)$ you can use this lemma:




          Let $x > y ge 0$ and $n in mathbbN$. Then $$x^1/n - y^1/n le (x-y)^1/n$$




          Proof:



          Let $t = left(fracyxright)^1/n in [0,1rangle$. Then $$(1-t)^n le 1-t le 1-t^n$$



          so $1-t le (1-t^n)^1/n$. Multiplying with $x^1/n$ gives the desired result.



          Now you should be able to finish with



          $$left|a^1/n - x^1/nright| le |a-x|^1/n$$



          Your attempt at $(2)$ looks good.






          share|cite|improve this answer




















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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            1
            down vote



            accepted










            For question 1, use the following geometric series identity:
            $$fracp^n - q^np - q = p^n-1 + p^n - 2 q + p^n - 3 q^2 + ldots + q^n - 1$$
            with $p = x^frac1n$ and $q = a^frac1n$.



            As for question 2, you're doing fine. There is what I think is an inconsequential typo under your first two inequalities, where you say $x > sqrt[3]frac1N$, where I think you may mean $x > a + sqrt[3]frac1N$, but it doesn't cause any problems down the road.



            It's hard for me to gauge exactly what is bothering you; I can only guess. However, what slightly bothers me is that the proof is presented slightly backwards. It's presented in the order of your working out, rather than in logical order.



            You're supposed to be able to take a number $N < 0$, and instantly produce a $delta > 0$ with the given property. So, your proof should probably take the form:




            Suppose $N < 0$, and let $delta = -sqrt[3]frac1N > 0$. Then $a - delta < x < a implies ldots implies frac1(x - a)^3 < N.$




            The working you've written is valuable working behind the scenes, but it's nicer to present the proof in the proper order.



            EDIT: In response to your comment, that's not too hard to do, and I think it is worth covering. Anything you feel is shaky is worth delving into. A proof is supposed to be convincing, and if it can't even convince the person who wrote it, then it's not working.



            To prove this, suppose that $a + sqrt[3]frac1N < x < a$. Then
            $$x - a > sqrt[3]frac1N.$$
            Then, because the function $x mapsto x^3$ is increasing, we have
            $$(x - a)^3 > left(sqrt[3]frac1Nright)^3 = frac1N.$$
            Now, $x < a$, so $x - a < 0$. For negative numbers, the map $x mapsto frac1x$ is decreasing, so
            $$frac1(x - a)^3 < N,$$
            which is what we wanted to prove.






            share|cite|improve this answer






















            • Thank you for your response. To clarify what exactly was bothering me about my proof: I feel like there could be some way to explain how we conclude that $frac1(x-a)^3 < N$ from the premise that $sqrt[3]frac1N + a < x < a$ for part (a), and likewise for part (b). Or is it unnecessary to explain this since it is just a simple fact from mere observation?
              – uznam
              Sep 9 at 13:36







            • 1




              @uznam Edited. Hope it helps.
              – Theo Bendit
              Sep 9 at 14:37














            up vote
            1
            down vote



            accepted










            For question 1, use the following geometric series identity:
            $$fracp^n - q^np - q = p^n-1 + p^n - 2 q + p^n - 3 q^2 + ldots + q^n - 1$$
            with $p = x^frac1n$ and $q = a^frac1n$.



            As for question 2, you're doing fine. There is what I think is an inconsequential typo under your first two inequalities, where you say $x > sqrt[3]frac1N$, where I think you may mean $x > a + sqrt[3]frac1N$, but it doesn't cause any problems down the road.



            It's hard for me to gauge exactly what is bothering you; I can only guess. However, what slightly bothers me is that the proof is presented slightly backwards. It's presented in the order of your working out, rather than in logical order.



            You're supposed to be able to take a number $N < 0$, and instantly produce a $delta > 0$ with the given property. So, your proof should probably take the form:




            Suppose $N < 0$, and let $delta = -sqrt[3]frac1N > 0$. Then $a - delta < x < a implies ldots implies frac1(x - a)^3 < N.$




            The working you've written is valuable working behind the scenes, but it's nicer to present the proof in the proper order.



            EDIT: In response to your comment, that's not too hard to do, and I think it is worth covering. Anything you feel is shaky is worth delving into. A proof is supposed to be convincing, and if it can't even convince the person who wrote it, then it's not working.



            To prove this, suppose that $a + sqrt[3]frac1N < x < a$. Then
            $$x - a > sqrt[3]frac1N.$$
            Then, because the function $x mapsto x^3$ is increasing, we have
            $$(x - a)^3 > left(sqrt[3]frac1Nright)^3 = frac1N.$$
            Now, $x < a$, so $x - a < 0$. For negative numbers, the map $x mapsto frac1x$ is decreasing, so
            $$frac1(x - a)^3 < N,$$
            which is what we wanted to prove.






            share|cite|improve this answer






















            • Thank you for your response. To clarify what exactly was bothering me about my proof: I feel like there could be some way to explain how we conclude that $frac1(x-a)^3 < N$ from the premise that $sqrt[3]frac1N + a < x < a$ for part (a), and likewise for part (b). Or is it unnecessary to explain this since it is just a simple fact from mere observation?
              – uznam
              Sep 9 at 13:36







            • 1




              @uznam Edited. Hope it helps.
              – Theo Bendit
              Sep 9 at 14:37












            up vote
            1
            down vote



            accepted







            up vote
            1
            down vote



            accepted






            For question 1, use the following geometric series identity:
            $$fracp^n - q^np - q = p^n-1 + p^n - 2 q + p^n - 3 q^2 + ldots + q^n - 1$$
            with $p = x^frac1n$ and $q = a^frac1n$.



            As for question 2, you're doing fine. There is what I think is an inconsequential typo under your first two inequalities, where you say $x > sqrt[3]frac1N$, where I think you may mean $x > a + sqrt[3]frac1N$, but it doesn't cause any problems down the road.



            It's hard for me to gauge exactly what is bothering you; I can only guess. However, what slightly bothers me is that the proof is presented slightly backwards. It's presented in the order of your working out, rather than in logical order.



            You're supposed to be able to take a number $N < 0$, and instantly produce a $delta > 0$ with the given property. So, your proof should probably take the form:




            Suppose $N < 0$, and let $delta = -sqrt[3]frac1N > 0$. Then $a - delta < x < a implies ldots implies frac1(x - a)^3 < N.$




            The working you've written is valuable working behind the scenes, but it's nicer to present the proof in the proper order.



            EDIT: In response to your comment, that's not too hard to do, and I think it is worth covering. Anything you feel is shaky is worth delving into. A proof is supposed to be convincing, and if it can't even convince the person who wrote it, then it's not working.



            To prove this, suppose that $a + sqrt[3]frac1N < x < a$. Then
            $$x - a > sqrt[3]frac1N.$$
            Then, because the function $x mapsto x^3$ is increasing, we have
            $$(x - a)^3 > left(sqrt[3]frac1Nright)^3 = frac1N.$$
            Now, $x < a$, so $x - a < 0$. For negative numbers, the map $x mapsto frac1x$ is decreasing, so
            $$frac1(x - a)^3 < N,$$
            which is what we wanted to prove.






            share|cite|improve this answer














            For question 1, use the following geometric series identity:
            $$fracp^n - q^np - q = p^n-1 + p^n - 2 q + p^n - 3 q^2 + ldots + q^n - 1$$
            with $p = x^frac1n$ and $q = a^frac1n$.



            As for question 2, you're doing fine. There is what I think is an inconsequential typo under your first two inequalities, where you say $x > sqrt[3]frac1N$, where I think you may mean $x > a + sqrt[3]frac1N$, but it doesn't cause any problems down the road.



            It's hard for me to gauge exactly what is bothering you; I can only guess. However, what slightly bothers me is that the proof is presented slightly backwards. It's presented in the order of your working out, rather than in logical order.



            You're supposed to be able to take a number $N < 0$, and instantly produce a $delta > 0$ with the given property. So, your proof should probably take the form:




            Suppose $N < 0$, and let $delta = -sqrt[3]frac1N > 0$. Then $a - delta < x < a implies ldots implies frac1(x - a)^3 < N.$




            The working you've written is valuable working behind the scenes, but it's nicer to present the proof in the proper order.



            EDIT: In response to your comment, that's not too hard to do, and I think it is worth covering. Anything you feel is shaky is worth delving into. A proof is supposed to be convincing, and if it can't even convince the person who wrote it, then it's not working.



            To prove this, suppose that $a + sqrt[3]frac1N < x < a$. Then
            $$x - a > sqrt[3]frac1N.$$
            Then, because the function $x mapsto x^3$ is increasing, we have
            $$(x - a)^3 > left(sqrt[3]frac1Nright)^3 = frac1N.$$
            Now, $x < a$, so $x - a < 0$. For negative numbers, the map $x mapsto frac1x$ is decreasing, so
            $$frac1(x - a)^3 < N,$$
            which is what we wanted to prove.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Sep 9 at 14:45

























            answered Sep 9 at 11:15









            Theo Bendit

            13.7k12045




            13.7k12045











            • Thank you for your response. To clarify what exactly was bothering me about my proof: I feel like there could be some way to explain how we conclude that $frac1(x-a)^3 < N$ from the premise that $sqrt[3]frac1N + a < x < a$ for part (a), and likewise for part (b). Or is it unnecessary to explain this since it is just a simple fact from mere observation?
              – uznam
              Sep 9 at 13:36







            • 1




              @uznam Edited. Hope it helps.
              – Theo Bendit
              Sep 9 at 14:37
















            • Thank you for your response. To clarify what exactly was bothering me about my proof: I feel like there could be some way to explain how we conclude that $frac1(x-a)^3 < N$ from the premise that $sqrt[3]frac1N + a < x < a$ for part (a), and likewise for part (b). Or is it unnecessary to explain this since it is just a simple fact from mere observation?
              – uznam
              Sep 9 at 13:36







            • 1




              @uznam Edited. Hope it helps.
              – Theo Bendit
              Sep 9 at 14:37















            Thank you for your response. To clarify what exactly was bothering me about my proof: I feel like there could be some way to explain how we conclude that $frac1(x-a)^3 < N$ from the premise that $sqrt[3]frac1N + a < x < a$ for part (a), and likewise for part (b). Or is it unnecessary to explain this since it is just a simple fact from mere observation?
            – uznam
            Sep 9 at 13:36





            Thank you for your response. To clarify what exactly was bothering me about my proof: I feel like there could be some way to explain how we conclude that $frac1(x-a)^3 < N$ from the premise that $sqrt[3]frac1N + a < x < a$ for part (a), and likewise for part (b). Or is it unnecessary to explain this since it is just a simple fact from mere observation?
            – uznam
            Sep 9 at 13:36





            1




            1




            @uznam Edited. Hope it helps.
            – Theo Bendit
            Sep 9 at 14:37




            @uznam Edited. Hope it helps.
            – Theo Bendit
            Sep 9 at 14:37










            up vote
            1
            down vote













            For $(1)$ you can use this lemma:




            Let $x > y ge 0$ and $n in mathbbN$. Then $$x^1/n - y^1/n le (x-y)^1/n$$




            Proof:



            Let $t = left(fracyxright)^1/n in [0,1rangle$. Then $$(1-t)^n le 1-t le 1-t^n$$



            so $1-t le (1-t^n)^1/n$. Multiplying with $x^1/n$ gives the desired result.



            Now you should be able to finish with



            $$left|a^1/n - x^1/nright| le |a-x|^1/n$$



            Your attempt at $(2)$ looks good.






            share|cite|improve this answer
























              up vote
              1
              down vote













              For $(1)$ you can use this lemma:




              Let $x > y ge 0$ and $n in mathbbN$. Then $$x^1/n - y^1/n le (x-y)^1/n$$




              Proof:



              Let $t = left(fracyxright)^1/n in [0,1rangle$. Then $$(1-t)^n le 1-t le 1-t^n$$



              so $1-t le (1-t^n)^1/n$. Multiplying with $x^1/n$ gives the desired result.



              Now you should be able to finish with



              $$left|a^1/n - x^1/nright| le |a-x|^1/n$$



              Your attempt at $(2)$ looks good.






              share|cite|improve this answer






















                up vote
                1
                down vote










                up vote
                1
                down vote









                For $(1)$ you can use this lemma:




                Let $x > y ge 0$ and $n in mathbbN$. Then $$x^1/n - y^1/n le (x-y)^1/n$$




                Proof:



                Let $t = left(fracyxright)^1/n in [0,1rangle$. Then $$(1-t)^n le 1-t le 1-t^n$$



                so $1-t le (1-t^n)^1/n$. Multiplying with $x^1/n$ gives the desired result.



                Now you should be able to finish with



                $$left|a^1/n - x^1/nright| le |a-x|^1/n$$



                Your attempt at $(2)$ looks good.






                share|cite|improve this answer












                For $(1)$ you can use this lemma:




                Let $x > y ge 0$ and $n in mathbbN$. Then $$x^1/n - y^1/n le (x-y)^1/n$$




                Proof:



                Let $t = left(fracyxright)^1/n in [0,1rangle$. Then $$(1-t)^n le 1-t le 1-t^n$$



                so $1-t le (1-t^n)^1/n$. Multiplying with $x^1/n$ gives the desired result.



                Now you should be able to finish with



                $$left|a^1/n - x^1/nright| le |a-x|^1/n$$



                Your attempt at $(2)$ looks good.







                share|cite|improve this answer












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                answered Sep 9 at 11:34









                mechanodroid

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