Solve $limlimits_{ntoinfty}sqrt[3]{n+sqrt{n}}-sqrt[3]{n}$











up vote
6
down vote

favorite
1












I am having great problems in solving this:



$$limlimits_{ntoinfty}sqrt[3]{n+sqrt{n}}-sqrt[3]{n}$$



I am trying to solve this for hours, no solution in sight. I tried so many ways on my paper here, which all lead to nonsense or to nowhere. I concluded that I have to use the third binomial formula here, so my next step would be:



$$a^3-b^3=(a-b)(a^2+ab+b^2)$$ so



$$a-b=frac{a^3-b^3}{a^2+ab+b^2}$$



I tried expanding it as well, which led to absolutely nothing. These are my writings to this:



enter image description here






limits roots







share|cite|improve this question




























    up vote
    6
    down vote

    favorite
    1












    I am having great problems in solving this:



    $$limlimits_{ntoinfty}sqrt[3]{n+sqrt{n}}-sqrt[3]{n}$$



    I am trying to solve this for hours, no solution in sight. I tried so many ways on my paper here, which all lead to nonsense or to nowhere. I concluded that I have to use the third binomial formula here, so my next step would be:



    $$a^3-b^3=(a-b)(a^2+ab+b^2)$$ so



    $$a-b=frac{a^3-b^3}{a^2+ab+b^2}$$



    I tried expanding it as well, which led to absolutely nothing. These are my writings to this:



    enter image description here






    limits roots







    share|cite|improve this question


























      up vote
      6
      down vote

      favorite
      1









      up vote
      6
      down vote

      favorite
      1






      1





      I am having great problems in solving this:



      $$limlimits_{ntoinfty}sqrt[3]{n+sqrt{n}}-sqrt[3]{n}$$



      I am trying to solve this for hours, no solution in sight. I tried so many ways on my paper here, which all lead to nonsense or to nowhere. I concluded that I have to use the third binomial formula here, so my next step would be:



      $$a^3-b^3=(a-b)(a^2+ab+b^2)$$ so



      $$a-b=frac{a^3-b^3}{a^2+ab+b^2}$$



      I tried expanding it as well, which led to absolutely nothing. These are my writings to this:



      enter image description here






      limits roots







      share|cite|improve this question















      I am having great problems in solving this:



      $$limlimits_{ntoinfty}sqrt[3]{n+sqrt{n}}-sqrt[3]{n}$$



      I am trying to solve this for hours, no solution in sight. I tried so many ways on my paper here, which all lead to nonsense or to nowhere. I concluded that I have to use the third binomial formula here, so my next step would be:



      $$a^3-b^3=(a-b)(a^2+ab+b^2)$$ so



      $$a-b=frac{a^3-b^3}{a^2+ab+b^2}$$



      I tried expanding it as well, which led to absolutely nothing. These are my writings to this:



      enter image description here






      limits roots




      limits roots






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 22 at 20:10









      Robert Howard

      1,9181822




      1,9181822










      asked Nov 22 at 19:28









      SacredScout

      487




      487






















          5 Answers
          5






          active

          oldest

          votes

















          up vote
          7
          down vote



          accepted










          $sqrt[3]{n+sqrt{n}}-sqrt[3]{n}=frac{n+sqrt{n}-n}{(n+sqrt{n})^{2/3}+(n+sqrt{n})^{1/3}n^{1/3}+n^{2/3}}leq frac{sqrt{n}}{n^{2/3}+n^{1/3}n^{1/3}+n^{2/3}}=frac{sqrt{n}}{3n^{2/3}}rightarrow 0$






          share|cite|improve this answer





















          • That is the clearest explanation I got, I can really understand everything there - which I can seldom say because my knowledge of maths is very limited (for now). I have chosen yours as the best, thank you! Keep your service up!
            – SacredScout
            Nov 22 at 20:02


















          up vote
          3
          down vote













          By first order binomial expansion $(1+x)^r=1+rx + o(x)$, we have



          $$sqrt[3]{n+sqrt{n}}=sqrt[3]{n}, left(1+frac1{sqrt n}right)^frac13=sqrt[3]{n}+frac{sqrt[3]{n}}{3sqrt n}+oleft(frac{sqrt[3]{n}}{sqrt n}right)=sqrt[3]{n}+frac{1}{3n^frac16}+oleft(frac{1}{n^frac16}right)$$



          therefore



          $$sqrt[3]{n+sqrt{n}}-sqrt[3]{n}=frac{1}{3n^frac16}+oleft(frac{1}{n^frac16}right)$$






          share|cite|improve this answer























          • May I ask why $(1+x)^r=1+rx+o(x)$ when $r$ is fractional?
            – William McGonagall
            Nov 27 at 7:08






          • 1




            @WilliamMcGonagall Yes of course! By binomial theorem $(1+x)^r=1+rx+r(r-1)^x^2+...=1+rx+o(x). Refer also to Binomial Series
            – gimusi
            Nov 27 at 7:14










          • Thanks for the pointer. The binomial theorem for fractional exponent looks mysterious to me.
            – William McGonagall
            Nov 27 at 7:26










          • @WilliamMcGonagall It can be derived by Taylor's expansion, are you aware about it?
            – gimusi
            Nov 27 at 7:37










          • Yes, but I wasn't familiar with the order of growth of $binom{r}{k}$ and so I didn't know what the radius of convergence of the series would be.
            – William McGonagall
            Nov 27 at 7:42


















          up vote
          3
          down vote













          Alternative approach where we do not use of the identity $$a^3-b^3=(a-b)(a^2+ab+b^2).$$



          We have that
          $$0< sqrt[3]{n+sqrt{n}}-sqrt[3]{n}=frac{sqrt[3]{n}}{sqrt{n}}cdotfrac{
          sqrt[3]{1+frac{1}{sqrt{n}}}-1}{frac{1}{sqrt{n}}}leq frac{1}{sqrt[6]{n}}cdot frac{1}{3}.$$
          where we used the Bernoulli's inequality
          $$(1+x)^rleq 1+rx$$
          with $r=1/3in(0,1)$ and $x=1/sqrt{n}$.



          Can you take it from here?



          P.S. By using this approach we are also able to find
          $$limlimits_{ntoinfty}left((n+sqrt{n})^r-n^rright)$$
          for any $r<1/2$. Note that if we replace $sqrt[3]{ }$ with say $sqrt[5]{ }$ then the "algebraic way" could be very annoying!






          share|cite|improve this answer























          • That looks really plausible and I could basically conclude that it converges to 0 with your term. But I have no Idea how I could rewrite the term in your manner, so that would sadly not help me in other tasks :(.... How did you rewrite it like that, by using which rules?
            – SacredScout
            Nov 22 at 19:37






          • 1




            Do you agree that $sqrt[3]{n}cdot ( sqrt[3]{1+frac{1}{sqrt{n}}}-1)=sqrt[3]{n+sqrt{n}}-sqrt[3]{n}$? Recall that $sqrt[3]{a}sqrt[3]{b}=sqrt[3]{ab}$.
            – Robert Z
            Nov 22 at 19:39












          • I must have been blind..... Got that! I still have to admit that I don't get the denominator.
            – SacredScout
            Nov 22 at 19:45










          • $sqrt{n}cdot frac{1}{sqrt{n}}=1$...
            – Robert Z
            Nov 22 at 19:46










          • It is much better now, but still really tough for me to understand. Would it be another questions I would have great problems with it, no doubt. I started studying 2 months ago, so eventually I will fully understand that as well. Thanks, Sir!
            – SacredScout
            Nov 27 at 15:23




















          up vote
          2
          down vote













          As you suggested,
          begin{align}
          sqrt[3]{n+sqrt n} - sqrt[3]{n} &= frac{n+sqrt n - n}{(n+sqrt n)^{2/3}+sqrt[3]{n(n+sqrt n)} + n^{2/3}}\
          &= frac{sqrt n}{(n+sqrt n)^{2/3}+sqrt[3]{n(n+sqrt n)} + n^{2/3}}.
          end{align}
          The fastest growing term is evidently $n^{2/3}$. Can you finish?






          share|cite|improve this answer




























            up vote
            0
            down vote













            Consider the function $f(x)=x^{1/3}$. By the mean value theorem there's a number $yin (n, n+sqrt n)$ such that
            $$
            f(n+sqrt n) - f(n) = f'(y)(n+sqrt n - n)= frac{y^{-2/3}}{3}sqrt n<n^{-2/3}sqrt n=n^{-1/6}to 0.
            $$






            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%2f3009543%2fsolve-lim-limits-n-to-infty-sqrt3n-sqrtn-sqrt3n%23new-answer', 'question_page');
              }
              );

              Post as a guest















              Required, but never shown

























              5 Answers
              5






              active

              oldest

              votes








              5 Answers
              5






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes








              up vote
              7
              down vote



              accepted










              $sqrt[3]{n+sqrt{n}}-sqrt[3]{n}=frac{n+sqrt{n}-n}{(n+sqrt{n})^{2/3}+(n+sqrt{n})^{1/3}n^{1/3}+n^{2/3}}leq frac{sqrt{n}}{n^{2/3}+n^{1/3}n^{1/3}+n^{2/3}}=frac{sqrt{n}}{3n^{2/3}}rightarrow 0$






              share|cite|improve this answer





















              • That is the clearest explanation I got, I can really understand everything there - which I can seldom say because my knowledge of maths is very limited (for now). I have chosen yours as the best, thank you! Keep your service up!
                – SacredScout
                Nov 22 at 20:02















              up vote
              7
              down vote



              accepted










              $sqrt[3]{n+sqrt{n}}-sqrt[3]{n}=frac{n+sqrt{n}-n}{(n+sqrt{n})^{2/3}+(n+sqrt{n})^{1/3}n^{1/3}+n^{2/3}}leq frac{sqrt{n}}{n^{2/3}+n^{1/3}n^{1/3}+n^{2/3}}=frac{sqrt{n}}{3n^{2/3}}rightarrow 0$






              share|cite|improve this answer





















              • That is the clearest explanation I got, I can really understand everything there - which I can seldom say because my knowledge of maths is very limited (for now). I have chosen yours as the best, thank you! Keep your service up!
                – SacredScout
                Nov 22 at 20:02













              up vote
              7
              down vote



              accepted







              up vote
              7
              down vote



              accepted






              $sqrt[3]{n+sqrt{n}}-sqrt[3]{n}=frac{n+sqrt{n}-n}{(n+sqrt{n})^{2/3}+(n+sqrt{n})^{1/3}n^{1/3}+n^{2/3}}leq frac{sqrt{n}}{n^{2/3}+n^{1/3}n^{1/3}+n^{2/3}}=frac{sqrt{n}}{3n^{2/3}}rightarrow 0$






              share|cite|improve this answer












              $sqrt[3]{n+sqrt{n}}-sqrt[3]{n}=frac{n+sqrt{n}-n}{(n+sqrt{n})^{2/3}+(n+sqrt{n})^{1/3}n^{1/3}+n^{2/3}}leq frac{sqrt{n}}{n^{2/3}+n^{1/3}n^{1/3}+n^{2/3}}=frac{sqrt{n}}{3n^{2/3}}rightarrow 0$







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Nov 22 at 19:36









              John_Wick

              1,199111




              1,199111












              • That is the clearest explanation I got, I can really understand everything there - which I can seldom say because my knowledge of maths is very limited (for now). I have chosen yours as the best, thank you! Keep your service up!
                – SacredScout
                Nov 22 at 20:02


















              • That is the clearest explanation I got, I can really understand everything there - which I can seldom say because my knowledge of maths is very limited (for now). I have chosen yours as the best, thank you! Keep your service up!
                – SacredScout
                Nov 22 at 20:02
















              That is the clearest explanation I got, I can really understand everything there - which I can seldom say because my knowledge of maths is very limited (for now). I have chosen yours as the best, thank you! Keep your service up!
              – SacredScout
              Nov 22 at 20:02




              That is the clearest explanation I got, I can really understand everything there - which I can seldom say because my knowledge of maths is very limited (for now). I have chosen yours as the best, thank you! Keep your service up!
              – SacredScout
              Nov 22 at 20:02










              up vote
              3
              down vote













              By first order binomial expansion $(1+x)^r=1+rx + o(x)$, we have



              $$sqrt[3]{n+sqrt{n}}=sqrt[3]{n}, left(1+frac1{sqrt n}right)^frac13=sqrt[3]{n}+frac{sqrt[3]{n}}{3sqrt n}+oleft(frac{sqrt[3]{n}}{sqrt n}right)=sqrt[3]{n}+frac{1}{3n^frac16}+oleft(frac{1}{n^frac16}right)$$



              therefore



              $$sqrt[3]{n+sqrt{n}}-sqrt[3]{n}=frac{1}{3n^frac16}+oleft(frac{1}{n^frac16}right)$$






              share|cite|improve this answer























              • May I ask why $(1+x)^r=1+rx+o(x)$ when $r$ is fractional?
                – William McGonagall
                Nov 27 at 7:08






              • 1




                @WilliamMcGonagall Yes of course! By binomial theorem $(1+x)^r=1+rx+r(r-1)^x^2+...=1+rx+o(x). Refer also to Binomial Series
                – gimusi
                Nov 27 at 7:14










              • Thanks for the pointer. The binomial theorem for fractional exponent looks mysterious to me.
                – William McGonagall
                Nov 27 at 7:26










              • @WilliamMcGonagall It can be derived by Taylor's expansion, are you aware about it?
                – gimusi
                Nov 27 at 7:37










              • Yes, but I wasn't familiar with the order of growth of $binom{r}{k}$ and so I didn't know what the radius of convergence of the series would be.
                – William McGonagall
                Nov 27 at 7:42















              up vote
              3
              down vote













              By first order binomial expansion $(1+x)^r=1+rx + o(x)$, we have



              $$sqrt[3]{n+sqrt{n}}=sqrt[3]{n}, left(1+frac1{sqrt n}right)^frac13=sqrt[3]{n}+frac{sqrt[3]{n}}{3sqrt n}+oleft(frac{sqrt[3]{n}}{sqrt n}right)=sqrt[3]{n}+frac{1}{3n^frac16}+oleft(frac{1}{n^frac16}right)$$



              therefore



              $$sqrt[3]{n+sqrt{n}}-sqrt[3]{n}=frac{1}{3n^frac16}+oleft(frac{1}{n^frac16}right)$$






              share|cite|improve this answer























              • May I ask why $(1+x)^r=1+rx+o(x)$ when $r$ is fractional?
                – William McGonagall
                Nov 27 at 7:08






              • 1




                @WilliamMcGonagall Yes of course! By binomial theorem $(1+x)^r=1+rx+r(r-1)^x^2+...=1+rx+o(x). Refer also to Binomial Series
                – gimusi
                Nov 27 at 7:14










              • Thanks for the pointer. The binomial theorem for fractional exponent looks mysterious to me.
                – William McGonagall
                Nov 27 at 7:26










              • @WilliamMcGonagall It can be derived by Taylor's expansion, are you aware about it?
                – gimusi
                Nov 27 at 7:37










              • Yes, but I wasn't familiar with the order of growth of $binom{r}{k}$ and so I didn't know what the radius of convergence of the series would be.
                – William McGonagall
                Nov 27 at 7:42













              up vote
              3
              down vote










              up vote
              3
              down vote









              By first order binomial expansion $(1+x)^r=1+rx + o(x)$, we have



              $$sqrt[3]{n+sqrt{n}}=sqrt[3]{n}, left(1+frac1{sqrt n}right)^frac13=sqrt[3]{n}+frac{sqrt[3]{n}}{3sqrt n}+oleft(frac{sqrt[3]{n}}{sqrt n}right)=sqrt[3]{n}+frac{1}{3n^frac16}+oleft(frac{1}{n^frac16}right)$$



              therefore



              $$sqrt[3]{n+sqrt{n}}-sqrt[3]{n}=frac{1}{3n^frac16}+oleft(frac{1}{n^frac16}right)$$






              share|cite|improve this answer














              By first order binomial expansion $(1+x)^r=1+rx + o(x)$, we have



              $$sqrt[3]{n+sqrt{n}}=sqrt[3]{n}, left(1+frac1{sqrt n}right)^frac13=sqrt[3]{n}+frac{sqrt[3]{n}}{3sqrt n}+oleft(frac{sqrt[3]{n}}{sqrt n}right)=sqrt[3]{n}+frac{1}{3n^frac16}+oleft(frac{1}{n^frac16}right)$$



              therefore



              $$sqrt[3]{n+sqrt{n}}-sqrt[3]{n}=frac{1}{3n^frac16}+oleft(frac{1}{n^frac16}right)$$







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Nov 22 at 23:45









              amWhy

              191k28224439




              191k28224439










              answered Nov 22 at 19:59









              gimusi

              92.8k94495




              92.8k94495












              • May I ask why $(1+x)^r=1+rx+o(x)$ when $r$ is fractional?
                – William McGonagall
                Nov 27 at 7:08






              • 1




                @WilliamMcGonagall Yes of course! By binomial theorem $(1+x)^r=1+rx+r(r-1)^x^2+...=1+rx+o(x). Refer also to Binomial Series
                – gimusi
                Nov 27 at 7:14










              • Thanks for the pointer. The binomial theorem for fractional exponent looks mysterious to me.
                – William McGonagall
                Nov 27 at 7:26










              • @WilliamMcGonagall It can be derived by Taylor's expansion, are you aware about it?
                – gimusi
                Nov 27 at 7:37










              • Yes, but I wasn't familiar with the order of growth of $binom{r}{k}$ and so I didn't know what the radius of convergence of the series would be.
                – William McGonagall
                Nov 27 at 7:42


















              • May I ask why $(1+x)^r=1+rx+o(x)$ when $r$ is fractional?
                – William McGonagall
                Nov 27 at 7:08






              • 1




                @WilliamMcGonagall Yes of course! By binomial theorem $(1+x)^r=1+rx+r(r-1)^x^2+...=1+rx+o(x). Refer also to Binomial Series
                – gimusi
                Nov 27 at 7:14










              • Thanks for the pointer. The binomial theorem for fractional exponent looks mysterious to me.
                – William McGonagall
                Nov 27 at 7:26










              • @WilliamMcGonagall It can be derived by Taylor's expansion, are you aware about it?
                – gimusi
                Nov 27 at 7:37










              • Yes, but I wasn't familiar with the order of growth of $binom{r}{k}$ and so I didn't know what the radius of convergence of the series would be.
                – William McGonagall
                Nov 27 at 7:42
















              May I ask why $(1+x)^r=1+rx+o(x)$ when $r$ is fractional?
              – William McGonagall
              Nov 27 at 7:08




              May I ask why $(1+x)^r=1+rx+o(x)$ when $r$ is fractional?
              – William McGonagall
              Nov 27 at 7:08




              1




              1




              @WilliamMcGonagall Yes of course! By binomial theorem $(1+x)^r=1+rx+r(r-1)^x^2+...=1+rx+o(x). Refer also to Binomial Series
              – gimusi
              Nov 27 at 7:14




              @WilliamMcGonagall Yes of course! By binomial theorem $(1+x)^r=1+rx+r(r-1)^x^2+...=1+rx+o(x). Refer also to Binomial Series
              – gimusi
              Nov 27 at 7:14












              Thanks for the pointer. The binomial theorem for fractional exponent looks mysterious to me.
              – William McGonagall
              Nov 27 at 7:26




              Thanks for the pointer. The binomial theorem for fractional exponent looks mysterious to me.
              – William McGonagall
              Nov 27 at 7:26












              @WilliamMcGonagall It can be derived by Taylor's expansion, are you aware about it?
              – gimusi
              Nov 27 at 7:37




              @WilliamMcGonagall It can be derived by Taylor's expansion, are you aware about it?
              – gimusi
              Nov 27 at 7:37












              Yes, but I wasn't familiar with the order of growth of $binom{r}{k}$ and so I didn't know what the radius of convergence of the series would be.
              – William McGonagall
              Nov 27 at 7:42




              Yes, but I wasn't familiar with the order of growth of $binom{r}{k}$ and so I didn't know what the radius of convergence of the series would be.
              – William McGonagall
              Nov 27 at 7:42










              up vote
              3
              down vote













              Alternative approach where we do not use of the identity $$a^3-b^3=(a-b)(a^2+ab+b^2).$$



              We have that
              $$0< sqrt[3]{n+sqrt{n}}-sqrt[3]{n}=frac{sqrt[3]{n}}{sqrt{n}}cdotfrac{
              sqrt[3]{1+frac{1}{sqrt{n}}}-1}{frac{1}{sqrt{n}}}leq frac{1}{sqrt[6]{n}}cdot frac{1}{3}.$$
              where we used the Bernoulli's inequality
              $$(1+x)^rleq 1+rx$$
              with $r=1/3in(0,1)$ and $x=1/sqrt{n}$.



              Can you take it from here?



              P.S. By using this approach we are also able to find
              $$limlimits_{ntoinfty}left((n+sqrt{n})^r-n^rright)$$
              for any $r<1/2$. Note that if we replace $sqrt[3]{ }$ with say $sqrt[5]{ }$ then the "algebraic way" could be very annoying!






              share|cite|improve this answer























              • That looks really plausible and I could basically conclude that it converges to 0 with your term. But I have no Idea how I could rewrite the term in your manner, so that would sadly not help me in other tasks :(.... How did you rewrite it like that, by using which rules?
                – SacredScout
                Nov 22 at 19:37






              • 1




                Do you agree that $sqrt[3]{n}cdot ( sqrt[3]{1+frac{1}{sqrt{n}}}-1)=sqrt[3]{n+sqrt{n}}-sqrt[3]{n}$? Recall that $sqrt[3]{a}sqrt[3]{b}=sqrt[3]{ab}$.
                – Robert Z
                Nov 22 at 19:39












              • I must have been blind..... Got that! I still have to admit that I don't get the denominator.
                – SacredScout
                Nov 22 at 19:45










              • $sqrt{n}cdot frac{1}{sqrt{n}}=1$...
                – Robert Z
                Nov 22 at 19:46










              • It is much better now, but still really tough for me to understand. Would it be another questions I would have great problems with it, no doubt. I started studying 2 months ago, so eventually I will fully understand that as well. Thanks, Sir!
                – SacredScout
                Nov 27 at 15:23

















              up vote
              3
              down vote













              Alternative approach where we do not use of the identity $$a^3-b^3=(a-b)(a^2+ab+b^2).$$



              We have that
              $$0< sqrt[3]{n+sqrt{n}}-sqrt[3]{n}=frac{sqrt[3]{n}}{sqrt{n}}cdotfrac{
              sqrt[3]{1+frac{1}{sqrt{n}}}-1}{frac{1}{sqrt{n}}}leq frac{1}{sqrt[6]{n}}cdot frac{1}{3}.$$
              where we used the Bernoulli's inequality
              $$(1+x)^rleq 1+rx$$
              with $r=1/3in(0,1)$ and $x=1/sqrt{n}$.



              Can you take it from here?



              P.S. By using this approach we are also able to find
              $$limlimits_{ntoinfty}left((n+sqrt{n})^r-n^rright)$$
              for any $r<1/2$. Note that if we replace $sqrt[3]{ }$ with say $sqrt[5]{ }$ then the "algebraic way" could be very annoying!






              share|cite|improve this answer























              • That looks really plausible and I could basically conclude that it converges to 0 with your term. But I have no Idea how I could rewrite the term in your manner, so that would sadly not help me in other tasks :(.... How did you rewrite it like that, by using which rules?
                – SacredScout
                Nov 22 at 19:37






              • 1




                Do you agree that $sqrt[3]{n}cdot ( sqrt[3]{1+frac{1}{sqrt{n}}}-1)=sqrt[3]{n+sqrt{n}}-sqrt[3]{n}$? Recall that $sqrt[3]{a}sqrt[3]{b}=sqrt[3]{ab}$.
                – Robert Z
                Nov 22 at 19:39












              • I must have been blind..... Got that! I still have to admit that I don't get the denominator.
                – SacredScout
                Nov 22 at 19:45










              • $sqrt{n}cdot frac{1}{sqrt{n}}=1$...
                – Robert Z
                Nov 22 at 19:46










              • It is much better now, but still really tough for me to understand. Would it be another questions I would have great problems with it, no doubt. I started studying 2 months ago, so eventually I will fully understand that as well. Thanks, Sir!
                – SacredScout
                Nov 27 at 15:23















              up vote
              3
              down vote










              up vote
              3
              down vote









              Alternative approach where we do not use of the identity $$a^3-b^3=(a-b)(a^2+ab+b^2).$$



              We have that
              $$0< sqrt[3]{n+sqrt{n}}-sqrt[3]{n}=frac{sqrt[3]{n}}{sqrt{n}}cdotfrac{
              sqrt[3]{1+frac{1}{sqrt{n}}}-1}{frac{1}{sqrt{n}}}leq frac{1}{sqrt[6]{n}}cdot frac{1}{3}.$$
              where we used the Bernoulli's inequality
              $$(1+x)^rleq 1+rx$$
              with $r=1/3in(0,1)$ and $x=1/sqrt{n}$.



              Can you take it from here?



              P.S. By using this approach we are also able to find
              $$limlimits_{ntoinfty}left((n+sqrt{n})^r-n^rright)$$
              for any $r<1/2$. Note that if we replace $sqrt[3]{ }$ with say $sqrt[5]{ }$ then the "algebraic way" could be very annoying!






              share|cite|improve this answer














              Alternative approach where we do not use of the identity $$a^3-b^3=(a-b)(a^2+ab+b^2).$$



              We have that
              $$0< sqrt[3]{n+sqrt{n}}-sqrt[3]{n}=frac{sqrt[3]{n}}{sqrt{n}}cdotfrac{
              sqrt[3]{1+frac{1}{sqrt{n}}}-1}{frac{1}{sqrt{n}}}leq frac{1}{sqrt[6]{n}}cdot frac{1}{3}.$$
              where we used the Bernoulli's inequality
              $$(1+x)^rleq 1+rx$$
              with $r=1/3in(0,1)$ and $x=1/sqrt{n}$.



              Can you take it from here?



              P.S. By using this approach we are also able to find
              $$limlimits_{ntoinfty}left((n+sqrt{n})^r-n^rright)$$
              for any $r<1/2$. Note that if we replace $sqrt[3]{ }$ with say $sqrt[5]{ }$ then the "algebraic way" could be very annoying!







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Nov 27 at 6:34

























              answered Nov 22 at 19:30









              Robert Z

              92.4k1058129




              92.4k1058129












              • That looks really plausible and I could basically conclude that it converges to 0 with your term. But I have no Idea how I could rewrite the term in your manner, so that would sadly not help me in other tasks :(.... How did you rewrite it like that, by using which rules?
                – SacredScout
                Nov 22 at 19:37






              • 1




                Do you agree that $sqrt[3]{n}cdot ( sqrt[3]{1+frac{1}{sqrt{n}}}-1)=sqrt[3]{n+sqrt{n}}-sqrt[3]{n}$? Recall that $sqrt[3]{a}sqrt[3]{b}=sqrt[3]{ab}$.
                – Robert Z
                Nov 22 at 19:39












              • I must have been blind..... Got that! I still have to admit that I don't get the denominator.
                – SacredScout
                Nov 22 at 19:45










              • $sqrt{n}cdot frac{1}{sqrt{n}}=1$...
                – Robert Z
                Nov 22 at 19:46










              • It is much better now, but still really tough for me to understand. Would it be another questions I would have great problems with it, no doubt. I started studying 2 months ago, so eventually I will fully understand that as well. Thanks, Sir!
                – SacredScout
                Nov 27 at 15:23




















              • That looks really plausible and I could basically conclude that it converges to 0 with your term. But I have no Idea how I could rewrite the term in your manner, so that would sadly not help me in other tasks :(.... How did you rewrite it like that, by using which rules?
                – SacredScout
                Nov 22 at 19:37






              • 1




                Do you agree that $sqrt[3]{n}cdot ( sqrt[3]{1+frac{1}{sqrt{n}}}-1)=sqrt[3]{n+sqrt{n}}-sqrt[3]{n}$? Recall that $sqrt[3]{a}sqrt[3]{b}=sqrt[3]{ab}$.
                – Robert Z
                Nov 22 at 19:39












              • I must have been blind..... Got that! I still have to admit that I don't get the denominator.
                – SacredScout
                Nov 22 at 19:45










              • $sqrt{n}cdot frac{1}{sqrt{n}}=1$...
                – Robert Z
                Nov 22 at 19:46










              • It is much better now, but still really tough for me to understand. Would it be another questions I would have great problems with it, no doubt. I started studying 2 months ago, so eventually I will fully understand that as well. Thanks, Sir!
                – SacredScout
                Nov 27 at 15:23


















              That looks really plausible and I could basically conclude that it converges to 0 with your term. But I have no Idea how I could rewrite the term in your manner, so that would sadly not help me in other tasks :(.... How did you rewrite it like that, by using which rules?
              – SacredScout
              Nov 22 at 19:37




              That looks really plausible and I could basically conclude that it converges to 0 with your term. But I have no Idea how I could rewrite the term in your manner, so that would sadly not help me in other tasks :(.... How did you rewrite it like that, by using which rules?
              – SacredScout
              Nov 22 at 19:37




              1




              1




              Do you agree that $sqrt[3]{n}cdot ( sqrt[3]{1+frac{1}{sqrt{n}}}-1)=sqrt[3]{n+sqrt{n}}-sqrt[3]{n}$? Recall that $sqrt[3]{a}sqrt[3]{b}=sqrt[3]{ab}$.
              – Robert Z
              Nov 22 at 19:39






              Do you agree that $sqrt[3]{n}cdot ( sqrt[3]{1+frac{1}{sqrt{n}}}-1)=sqrt[3]{n+sqrt{n}}-sqrt[3]{n}$? Recall that $sqrt[3]{a}sqrt[3]{b}=sqrt[3]{ab}$.
              – Robert Z
              Nov 22 at 19:39














              I must have been blind..... Got that! I still have to admit that I don't get the denominator.
              – SacredScout
              Nov 22 at 19:45




              I must have been blind..... Got that! I still have to admit that I don't get the denominator.
              – SacredScout
              Nov 22 at 19:45












              $sqrt{n}cdot frac{1}{sqrt{n}}=1$...
              – Robert Z
              Nov 22 at 19:46




              $sqrt{n}cdot frac{1}{sqrt{n}}=1$...
              – Robert Z
              Nov 22 at 19:46












              It is much better now, but still really tough for me to understand. Would it be another questions I would have great problems with it, no doubt. I started studying 2 months ago, so eventually I will fully understand that as well. Thanks, Sir!
              – SacredScout
              Nov 27 at 15:23






              It is much better now, but still really tough for me to understand. Would it be another questions I would have great problems with it, no doubt. I started studying 2 months ago, so eventually I will fully understand that as well. Thanks, Sir!
              – SacredScout
              Nov 27 at 15:23












              up vote
              2
              down vote













              As you suggested,
              begin{align}
              sqrt[3]{n+sqrt n} - sqrt[3]{n} &= frac{n+sqrt n - n}{(n+sqrt n)^{2/3}+sqrt[3]{n(n+sqrt n)} + n^{2/3}}\
              &= frac{sqrt n}{(n+sqrt n)^{2/3}+sqrt[3]{n(n+sqrt n)} + n^{2/3}}.
              end{align}
              The fastest growing term is evidently $n^{2/3}$. Can you finish?






              share|cite|improve this answer

























                up vote
                2
                down vote













                As you suggested,
                begin{align}
                sqrt[3]{n+sqrt n} - sqrt[3]{n} &= frac{n+sqrt n - n}{(n+sqrt n)^{2/3}+sqrt[3]{n(n+sqrt n)} + n^{2/3}}\
                &= frac{sqrt n}{(n+sqrt n)^{2/3}+sqrt[3]{n(n+sqrt n)} + n^{2/3}}.
                end{align}
                The fastest growing term is evidently $n^{2/3}$. Can you finish?






                share|cite|improve this answer























                  up vote
                  2
                  down vote










                  up vote
                  2
                  down vote









                  As you suggested,
                  begin{align}
                  sqrt[3]{n+sqrt n} - sqrt[3]{n} &= frac{n+sqrt n - n}{(n+sqrt n)^{2/3}+sqrt[3]{n(n+sqrt n)} + n^{2/3}}\
                  &= frac{sqrt n}{(n+sqrt n)^{2/3}+sqrt[3]{n(n+sqrt n)} + n^{2/3}}.
                  end{align}
                  The fastest growing term is evidently $n^{2/3}$. Can you finish?






                  share|cite|improve this answer












                  As you suggested,
                  begin{align}
                  sqrt[3]{n+sqrt n} - sqrt[3]{n} &= frac{n+sqrt n - n}{(n+sqrt n)^{2/3}+sqrt[3]{n(n+sqrt n)} + n^{2/3}}\
                  &= frac{sqrt n}{(n+sqrt n)^{2/3}+sqrt[3]{n(n+sqrt n)} + n^{2/3}}.
                  end{align}
                  The fastest growing term is evidently $n^{2/3}$. Can you finish?







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 22 at 19:31









                  MisterRiemann

                  5,7131624




                  5,7131624






















                      up vote
                      0
                      down vote













                      Consider the function $f(x)=x^{1/3}$. By the mean value theorem there's a number $yin (n, n+sqrt n)$ such that
                      $$
                      f(n+sqrt n) - f(n) = f'(y)(n+sqrt n - n)= frac{y^{-2/3}}{3}sqrt n<n^{-2/3}sqrt n=n^{-1/6}to 0.
                      $$






                      share|cite|improve this answer

























                        up vote
                        0
                        down vote













                        Consider the function $f(x)=x^{1/3}$. By the mean value theorem there's a number $yin (n, n+sqrt n)$ such that
                        $$
                        f(n+sqrt n) - f(n) = f'(y)(n+sqrt n - n)= frac{y^{-2/3}}{3}sqrt n<n^{-2/3}sqrt n=n^{-1/6}to 0.
                        $$






                        share|cite|improve this answer























                          up vote
                          0
                          down vote










                          up vote
                          0
                          down vote









                          Consider the function $f(x)=x^{1/3}$. By the mean value theorem there's a number $yin (n, n+sqrt n)$ such that
                          $$
                          f(n+sqrt n) - f(n) = f'(y)(n+sqrt n - n)= frac{y^{-2/3}}{3}sqrt n<n^{-2/3}sqrt n=n^{-1/6}to 0.
                          $$






                          share|cite|improve this answer












                          Consider the function $f(x)=x^{1/3}$. By the mean value theorem there's a number $yin (n, n+sqrt n)$ such that
                          $$
                          f(n+sqrt n) - f(n) = f'(y)(n+sqrt n - n)= frac{y^{-2/3}}{3}sqrt n<n^{-2/3}sqrt n=n^{-1/6}to 0.
                          $$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 24 at 8:21









                          User

                          483514




                          483514






























                              draft saved

                              draft discarded




















































                              Thanks for contributing an answer to Mathematics Stack Exchange!


                              • Please be sure to answer the question. Provide details and share your research!

                              But avoid …



                              • Asking for help, clarification, or responding to other answers.

                              • Making statements based on opinion; back them up with references or personal experience.


                              Use MathJax to format equations. MathJax reference.


                              To learn more, see our tips on writing great answers.





                              Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                              Please pay close attention to the following guidance:


                              • Please be sure to answer the question. Provide details and share your research!

                              But avoid …



                              • Asking for help, clarification, or responding to other answers.

                              • Making statements based on opinion; back them up with references or personal experience.


                              To learn more, see our tips on writing great answers.




                              draft saved


                              draft discarded














                              StackExchange.ready(
                              function () {
                              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3009543%2fsolve-lim-limits-n-to-infty-sqrt3n-sqrtn-sqrt3n%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

                              Ellipse (mathématiques)

                              Image
                              Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
                              Read more

                              Quarter-circle Tiles

                              Image
                              up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
                              Read more

                              Mont Emei

                              Image
                              Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
                              Read more