Evaluting $lim_{n to infty } sqrt[n]{25n+n^3}$











up vote
2
down vote

favorite
3












$lim limits_{n to infty } sqrt[n]{25n+n^3}$



$1leftarrowsqrt[n]{25n}lesqrt[n]{25n+n^3}lesqrt[n]{25n^3+n^3}le sqrt[n]{26n^3}=sqrt[n]{26}cdotsqrt[n]{n^3}to1cdot1=1$



But is it obvious that $ forall _{kin Bbb N} (sqrt[n]n)^k to1 ?$



Since I know $(sqrt[n]n) to1$.



Is it enought to say $(sqrt[n]n)^kto1^k? $






real-analysis limits limits-without-lhopital







share|cite|improve this question
























  • $n^{3/n}=(n/3)^{3/n}cdot 3^{3/n}to 1$.
    – user254433
    yesterday






  • 1




    You can use the basic result: if $a_nto a$ and $b_nto b$ then $a_nb_n to ab$ to get $sqrt[n]{n}^k to 1^k = 1$. This works for integer $k$ (induction). In general you can use continuity of the power-function.
    – Winther
    yesterday












  • These answers actually show that if $p(n)$ is a polynomial in $n$, then $(p(n))^{1/n} to 1$.
    – JavaMan
    yesterday















up vote
2
down vote

favorite
3












$lim limits_{n to infty } sqrt[n]{25n+n^3}$



$1leftarrowsqrt[n]{25n}lesqrt[n]{25n+n^3}lesqrt[n]{25n^3+n^3}le sqrt[n]{26n^3}=sqrt[n]{26}cdotsqrt[n]{n^3}to1cdot1=1$



But is it obvious that $ forall _{kin Bbb N} (sqrt[n]n)^k to1 ?$



Since I know $(sqrt[n]n) to1$.



Is it enought to say $(sqrt[n]n)^kto1^k? $






real-analysis limits limits-without-lhopital







share|cite|improve this question
























  • $n^{3/n}=(n/3)^{3/n}cdot 3^{3/n}to 1$.
    – user254433
    yesterday






  • 1




    You can use the basic result: if $a_nto a$ and $b_nto b$ then $a_nb_n to ab$ to get $sqrt[n]{n}^k to 1^k = 1$. This works for integer $k$ (induction). In general you can use continuity of the power-function.
    – Winther
    yesterday












  • These answers actually show that if $p(n)$ is a polynomial in $n$, then $(p(n))^{1/n} to 1$.
    – JavaMan
    yesterday













up vote
2
down vote

favorite
3









up vote
2
down vote

favorite
3






3





$lim limits_{n to infty } sqrt[n]{25n+n^3}$



$1leftarrowsqrt[n]{25n}lesqrt[n]{25n+n^3}lesqrt[n]{25n^3+n^3}le sqrt[n]{26n^3}=sqrt[n]{26}cdotsqrt[n]{n^3}to1cdot1=1$



But is it obvious that $ forall _{kin Bbb N} (sqrt[n]n)^k to1 ?$



Since I know $(sqrt[n]n) to1$.



Is it enought to say $(sqrt[n]n)^kto1^k? $






real-analysis limits limits-without-lhopital







share|cite|improve this question















$lim limits_{n to infty } sqrt[n]{25n+n^3}$



$1leftarrowsqrt[n]{25n}lesqrt[n]{25n+n^3}lesqrt[n]{25n^3+n^3}le sqrt[n]{26n^3}=sqrt[n]{26}cdotsqrt[n]{n^3}to1cdot1=1$



But is it obvious that $ forall _{kin Bbb N} (sqrt[n]n)^k to1 ?$



Since I know $(sqrt[n]n) to1$.



Is it enought to say $(sqrt[n]n)^kto1^k? $






real-analysis limits limits-without-lhopital




real-analysis limits limits-without-lhopital






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited yesterday









amWhy

191k27223437




191k27223437










asked yesterday









matematiccc

1035




1035












  • $n^{3/n}=(n/3)^{3/n}cdot 3^{3/n}to 1$.
    – user254433
    yesterday






  • 1




    You can use the basic result: if $a_nto a$ and $b_nto b$ then $a_nb_n to ab$ to get $sqrt[n]{n}^k to 1^k = 1$. This works for integer $k$ (induction). In general you can use continuity of the power-function.
    – Winther
    yesterday












  • These answers actually show that if $p(n)$ is a polynomial in $n$, then $(p(n))^{1/n} to 1$.
    – JavaMan
    yesterday


















  • $n^{3/n}=(n/3)^{3/n}cdot 3^{3/n}to 1$.
    – user254433
    yesterday






  • 1




    You can use the basic result: if $a_nto a$ and $b_nto b$ then $a_nb_n to ab$ to get $sqrt[n]{n}^k to 1^k = 1$. This works for integer $k$ (induction). In general you can use continuity of the power-function.
    – Winther
    yesterday












  • These answers actually show that if $p(n)$ is a polynomial in $n$, then $(p(n))^{1/n} to 1$.
    – JavaMan
    yesterday
















$n^{3/n}=(n/3)^{3/n}cdot 3^{3/n}to 1$.
– user254433
yesterday




$n^{3/n}=(n/3)^{3/n}cdot 3^{3/n}to 1$.
– user254433
yesterday




1




1




You can use the basic result: if $a_nto a$ and $b_nto b$ then $a_nb_n to ab$ to get $sqrt[n]{n}^k to 1^k = 1$. This works for integer $k$ (induction). In general you can use continuity of the power-function.
– Winther
yesterday






You can use the basic result: if $a_nto a$ and $b_nto b$ then $a_nb_n to ab$ to get $sqrt[n]{n}^k to 1^k = 1$. This works for integer $k$ (induction). In general you can use continuity of the power-function.
– Winther
yesterday














These answers actually show that if $p(n)$ is a polynomial in $n$, then $(p(n))^{1/n} to 1$.
– JavaMan
yesterday




These answers actually show that if $p(n)$ is a polynomial in $n$, then $(p(n))^{1/n} to 1$.
– JavaMan
yesterday










3 Answers
3






active

oldest

votes

















up vote
3
down vote



accepted










Yes your idea is correct indeed we have that eventually



$$sqrt[n]{25n}le sqrt[n]{25n+n^3}le sqrt[n]{2n^3}$$



and




  • $sqrt[n]{25n}=sqrt[n]{25}sqrt[n]{n}to 1$

  • $sqrt[n]{2n^3}=(sqrt[n]{2})^3(sqrt[n]{n})^3 to 1$






share|cite|improve this answer




























    up vote
    0
    down vote













    use $limsup_{n rightarrow infty} |a_n|^{1/n}$ to help you find the answer.






    share|cite|improve this answer










    New contributor




    pfmr1995 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.

























      up vote
      0
      down vote













      You could also have good approximations for finite values of $n$.



      $$a_n=sqrt[n]{kn+n^3}implies log(a_n)=frac 1 n log(kn+n^3)=frac 1 n left(3 log(n)+log left(1+frac{k}{n^2}right) right)$$ Now, using Taylor expansions for large values of $n$
      $$log(a_n)=frac{3 log
      left({n}right)}{n}+frac{k}{n^3}+Oleft(frac{1}{n^5}right)$$       Continuing with Taylor
      $$a_n=e^{log(a_n)}=1+frac{3 log left({n}right)}{n}+frac{9 log
      ^2left({n}right)}{2 n^2}+frac{6 k+27 log
      ^3left({n}right)}{6 n^3}+Oleft(frac{1}{n^4}right)$$       which, for sure, shows the limit and how "far" is $k$ appearing in the approximation.






      share|cite|improve this answer





















        Your Answer





        StackExchange.ifUsing("editor", function () {
        return StackExchange.using("mathjaxEditing", function () {
        StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
        StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
        });
        });
        }, "mathjax-editing");

        StackExchange.ready(function() {
        var channelOptions = {
        tags: "".split(" "),
        id: "69"
        };
        initTagRenderer("".split(" "), "".split(" "), channelOptions);

        StackExchange.using("externalEditor", function() {
        // Have to fire editor after snippets, if snippets enabled
        if (StackExchange.settings.snippets.snippetsEnabled) {
        StackExchange.using("snippets", function() {
        createEditor();
        });
        }
        else {
        createEditor();
        }
        });

        function createEditor() {
        StackExchange.prepareEditor({
        heartbeatType: 'answer',
        convertImagesToLinks: true,
        noModals: true,
        showLowRepImageUploadWarning: true,
        reputationToPostImages: 10,
        bindNavPrevention: true,
        postfix: "",
        imageUploader: {
        brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
        contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
        allowUrls: true
        },
        noCode: true, onDemand: true,
        discardSelector: ".discard-answer"
        ,immediatelyShowMarkdownHelp:true
        });


        }
        });














         

        draft saved


        draft discarded


















        StackExchange.ready(
        function () {
        StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3004268%2fevaluting-lim-n-to-infty-sqrtn25nn3%23new-answer', 'question_page');
        }
        );

        Post as a guest















        Required, but never shown

























        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes








        up vote
        3
        down vote



        accepted










        Yes your idea is correct indeed we have that eventually



        $$sqrt[n]{25n}le sqrt[n]{25n+n^3}le sqrt[n]{2n^3}$$



        and




        • $sqrt[n]{25n}=sqrt[n]{25}sqrt[n]{n}to 1$

        • $sqrt[n]{2n^3}=(sqrt[n]{2})^3(sqrt[n]{n})^3 to 1$






        share|cite|improve this answer

























          up vote
          3
          down vote



          accepted










          Yes your idea is correct indeed we have that eventually



          $$sqrt[n]{25n}le sqrt[n]{25n+n^3}le sqrt[n]{2n^3}$$



          and




          • $sqrt[n]{25n}=sqrt[n]{25}sqrt[n]{n}to 1$

          • $sqrt[n]{2n^3}=(sqrt[n]{2})^3(sqrt[n]{n})^3 to 1$






          share|cite|improve this answer























            up vote
            3
            down vote



            accepted







            up vote
            3
            down vote



            accepted






            Yes your idea is correct indeed we have that eventually



            $$sqrt[n]{25n}le sqrt[n]{25n+n^3}le sqrt[n]{2n^3}$$



            and




            • $sqrt[n]{25n}=sqrt[n]{25}sqrt[n]{n}to 1$

            • $sqrt[n]{2n^3}=(sqrt[n]{2})^3(sqrt[n]{n})^3 to 1$






            share|cite|improve this answer












            Yes your idea is correct indeed we have that eventually



            $$sqrt[n]{25n}le sqrt[n]{25n+n^3}le sqrt[n]{2n^3}$$



            and




            • $sqrt[n]{25n}=sqrt[n]{25}sqrt[n]{n}to 1$

            • $sqrt[n]{2n^3}=(sqrt[n]{2})^3(sqrt[n]{n})^3 to 1$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered yesterday









            gimusi

            85.5k74294




            85.5k74294






















                up vote
                0
                down vote













                use $limsup_{n rightarrow infty} |a_n|^{1/n}$ to help you find the answer.






                share|cite|improve this answer










                New contributor




                pfmr1995 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.






















                  up vote
                  0
                  down vote













                  use $limsup_{n rightarrow infty} |a_n|^{1/n}$ to help you find the answer.






                  share|cite|improve this answer










                  New contributor




                  pfmr1995 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.




















                    up vote
                    0
                    down vote










                    up vote
                    0
                    down vote









                    use $limsup_{n rightarrow infty} |a_n|^{1/n}$ to help you find the answer.






                    share|cite|improve this answer










                    New contributor




                    pfmr1995 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.









                    use $limsup_{n rightarrow infty} |a_n|^{1/n}$ to help you find the answer.







                    share|cite|improve this answer










                    New contributor




                    pfmr1995 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 answer



                    share|cite|improve this answer








                    edited yesterday









                    Tianlalu

                    2,549632




                    2,549632






                    New contributor




                    pfmr1995 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.









                    answered yesterday









                    pfmr1995

                    113




                    113




                    New contributor




                    pfmr1995 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.





                    New contributor





                    pfmr1995 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.






                    pfmr1995 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.






















                        up vote
                        0
                        down vote













                        You could also have good approximations for finite values of $n$.



                        $$a_n=sqrt[n]{kn+n^3}implies log(a_n)=frac 1 n log(kn+n^3)=frac 1 n left(3 log(n)+log left(1+frac{k}{n^2}right) right)$$ Now, using Taylor expansions for large values of $n$
                        $$log(a_n)=frac{3 log
                        left({n}right)}{n}+frac{k}{n^3}+Oleft(frac{1}{n^5}right)$$                         Continuing with Taylor
                        $$a_n=e^{log(a_n)}=1+frac{3 log left({n}right)}{n}+frac{9 log
                        ^2left({n}right)}{2 n^2}+frac{6 k+27 log
                        ^3left({n}right)}{6 n^3}+Oleft(frac{1}{n^4}right)$$                         which, for sure, shows the limit and how "far" is $k$ appearing in the approximation.






                        share|cite|improve this answer

























                          up vote
                          0
                          down vote













                          You could also have good approximations for finite values of $n$.



                          $$a_n=sqrt[n]{kn+n^3}implies log(a_n)=frac 1 n log(kn+n^3)=frac 1 n left(3 log(n)+log left(1+frac{k}{n^2}right) right)$$ Now, using Taylor expansions for large values of $n$
                          $$log(a_n)=frac{3 log
                          left({n}right)}{n}+frac{k}{n^3}+Oleft(frac{1}{n^5}right)$$                           Continuing with Taylor
                          $$a_n=e^{log(a_n)}=1+frac{3 log left({n}right)}{n}+frac{9 log
                          ^2left({n}right)}{2 n^2}+frac{6 k+27 log
                          ^3left({n}right)}{6 n^3}+Oleft(frac{1}{n^4}right)$$                           which, for sure, shows the limit and how "far" is $k$ appearing in the approximation.






                          share|cite|improve this answer























                            up vote
                            0
                            down vote










                            up vote
                            0
                            down vote









                            You could also have good approximations for finite values of $n$.



                            $$a_n=sqrt[n]{kn+n^3}implies log(a_n)=frac 1 n log(kn+n^3)=frac 1 n left(3 log(n)+log left(1+frac{k}{n^2}right) right)$$ Now, using Taylor expansions for large values of $n$
                            $$log(a_n)=frac{3 log
                            left({n}right)}{n}+frac{k}{n^3}+Oleft(frac{1}{n^5}right)$$                             Continuing with Taylor
                            $$a_n=e^{log(a_n)}=1+frac{3 log left({n}right)}{n}+frac{9 log
                            ^2left({n}right)}{2 n^2}+frac{6 k+27 log
                            ^3left({n}right)}{6 n^3}+Oleft(frac{1}{n^4}right)$$                             which, for sure, shows the limit and how "far" is $k$ appearing in the approximation.






                            share|cite|improve this answer












                            You could also have good approximations for finite values of $n$.



                            $$a_n=sqrt[n]{kn+n^3}implies log(a_n)=frac 1 n log(kn+n^3)=frac 1 n left(3 log(n)+log left(1+frac{k}{n^2}right) right)$$ Now, using Taylor expansions for large values of $n$
                            $$log(a_n)=frac{3 log
                            left({n}right)}{n}+frac{k}{n^3}+Oleft(frac{1}{n^5}right)$$                             Continuing with Taylor
                            $$a_n=e^{log(a_n)}=1+frac{3 log left({n}right)}{n}+frac{9 log
                            ^2left({n}right)}{2 n^2}+frac{6 k+27 log
                            ^3left({n}right)}{6 n^3}+Oleft(frac{1}{n^4}right)$$                             which, for sure, shows the limit and how "far" is $k$ appearing in the approximation.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 23 hours ago









                            Claude Leibovici

                            116k1156131




                            116k1156131






























                                 

                                draft saved


                                draft discarded



















































                                 


                                draft saved


                                draft discarded














                                StackExchange.ready(
                                function () {
                                StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3004268%2fevaluting-lim-n-to-infty-sqrtn25nn3%23new-answer', 'question_page');
                                }
                                );

                                Post as a guest















                                Required, but never shown





















































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown

































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown







                                Popular posts from this blog

                                'app-layout' is not a known element: how to share Component with different Modules

                                Image
                                0 My application was working fine until I decided to go with lazy loading: So, my shared component looks like this: import { Component, Renderer2 } from '@angular/core'; export interface FormModel { captcha?: string; } @Component({ selector: 'app-layout', templateUrl: './app-layout.component.html', styleUrls: ['./app-layout.css'] }) export class FullLayoutComponent { } app-layout.component.ts I am using this component in my Another component say school-home.component.html like this: <app-layout> </app-layout> and it was working fine until I created a Module school-home.module.ts like this: import { NgModule } from '@angular/core'; import { CommonModule } from '@angular/common'; import { SchoolhomeRoutingModule ...
                                Read more

                                android studio warns about leanback feature tag usage required on manifest while using Unity exported app?

                                Image
                                .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 This is a simple project showing cubes on Unity, exported to android project and imported on Android studio. On maifest file it showed following warning: Expecting <uses-feature android:name="android.software.leanback" android:required="false" /> tag I have no information about it. And also on default emptyview android project does not use/mention this feature. I know it is required by TV's and such but why its not added by default? Why is this? And Should I concern about it? android unity3d ...
                                Read more

                                WPF add header to Image with URL pettitions [duplicate]

                                Image
                                up vote 0 down vote favorite This question already has an answer here: Image source URI requiring authorization 3 answers Add header to all* HTTP requests on a machine 1 answer first of all sorry for my english. Recently I added some security in the backend of my project, and now I need to send a validation code(apiKey) inside the headers to retrieve the images. I have a problem with ALL images in my project that I had attached in my UI. For example, before I had: <Image Height="90" Source="{Binding Video.Thumbnail}"/> Now I need to change my code to create a CUSTOM petition add the apiKey in the header:...
                                Read more