Limit $ lim_{ntoinfty} left(frac{3n^2-n+1}{2n^2+n+1}right)^{large frac{n^2}{1-n}}$












3












$begingroup$



I'm required to compute the limit of the following sequence:
$$ lim_{ntoinfty} left(frac{3n^2-n+1}{2n^2+n+1}right)^{large frac{n^2}{1-n}}$$




I'm not sure as to how to approach it. I've tried tackling it in a few ways, but none of them led me to a form I know how to tackle.



Let $a_n$ denote the base.



$$ lim_{ntoinfty}(a_n)^{frac{n^2}{1-n}}= left(lim_{ntoinfty}(a_n)^{large frac{1}{1-n}}right)^{n^2}=lim_{ntoinfty} left(sqrt[1-n]{a_n}right)^{n^2}=1^{infty} $$



From there though I don't know how to compute:
$$e^{lim_{ntoinfty} left(sqrt[n-1]{a_n}n^2right)}$$



I tried using the ln function though I couldn't get any further. I'd be glad for help :)






real-analysis calculus limits







share|cite|improve this question











$endgroup$

















    3












    $begingroup$



    I'm required to compute the limit of the following sequence:
    $$ lim_{ntoinfty} left(frac{3n^2-n+1}{2n^2+n+1}right)^{large frac{n^2}{1-n}}$$




    I'm not sure as to how to approach it. I've tried tackling it in a few ways, but none of them led me to a form I know how to tackle.



    Let $a_n$ denote the base.



    $$ lim_{ntoinfty}(a_n)^{frac{n^2}{1-n}}= left(lim_{ntoinfty}(a_n)^{large frac{1}{1-n}}right)^{n^2}=lim_{ntoinfty} left(sqrt[1-n]{a_n}right)^{n^2}=1^{infty} $$



    From there though I don't know how to compute:
    $$e^{lim_{ntoinfty} left(sqrt[n-1]{a_n}n^2right)}$$



    I tried using the ln function though I couldn't get any further. I'd be glad for help :)






    real-analysis calculus limits







    share|cite|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$



      I'm required to compute the limit of the following sequence:
      $$ lim_{ntoinfty} left(frac{3n^2-n+1}{2n^2+n+1}right)^{large frac{n^2}{1-n}}$$




      I'm not sure as to how to approach it. I've tried tackling it in a few ways, but none of them led me to a form I know how to tackle.



      Let $a_n$ denote the base.



      $$ lim_{ntoinfty}(a_n)^{frac{n^2}{1-n}}= left(lim_{ntoinfty}(a_n)^{large frac{1}{1-n}}right)^{n^2}=lim_{ntoinfty} left(sqrt[1-n]{a_n}right)^{n^2}=1^{infty} $$



      From there though I don't know how to compute:
      $$e^{lim_{ntoinfty} left(sqrt[n-1]{a_n}n^2right)}$$



      I tried using the ln function though I couldn't get any further. I'd be glad for help :)






      real-analysis calculus limits







      share|cite|improve this question











      $endgroup$





      I'm required to compute the limit of the following sequence:
      $$ lim_{ntoinfty} left(frac{3n^2-n+1}{2n^2+n+1}right)^{large frac{n^2}{1-n}}$$




      I'm not sure as to how to approach it. I've tried tackling it in a few ways, but none of them led me to a form I know how to tackle.



      Let $a_n$ denote the base.



      $$ lim_{ntoinfty}(a_n)^{frac{n^2}{1-n}}= left(lim_{ntoinfty}(a_n)^{large frac{1}{1-n}}right)^{n^2}=lim_{ntoinfty} left(sqrt[1-n]{a_n}right)^{n^2}=1^{infty} $$



      From there though I don't know how to compute:
      $$e^{lim_{ntoinfty} left(sqrt[n-1]{a_n}n^2right)}$$



      I tried using the ln function though I couldn't get any further. I'd be glad for help :)






      real-analysis calculus limits




      real-analysis calculus limits






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 5 '18 at 13:26









      Arjang

      5,59062363




      5,59062363










      asked Dec 5 '18 at 10:47









      MosheMoshe

      396




      396






















          3 Answers
          3






          active

          oldest

          votes


















          4












          $begingroup$

          HINT



          $$left(frac{3n^2-n+1}{2n^2+n+1}right)^{frac{n^2}{1-n}}sim ccdot left(frac32right)^{-n}$$



          indeed



          $$left(frac{3n^2-n+1}{2n^2+n+1}right)^{frac{n^2}{1-n}}=left(frac32right)^{-n}left(frac{2n^2-frac23n+frac23}{2n^2+n+1}right)^{frac{n^2}{1-n}}$$



          and



          $$left(frac{2n^2-frac23n+frac23}{2n^2+n+1}right)^{frac{n^2}{1-n}}=left[left(1-frac{frac53n+frac13}{2n^2+n+1}right)^{frac{2n^2+n+1}{frac53n+frac13}}right]^{frac{frac53n^3+frac13n^2}{(2n^2+n+1)(1-n)}}toleft(frac1eright)^{-frac56}$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I observed that, it leads to 0 but subtituing larger values in the original form gives me other value
            $endgroup$
            – Moshe
            Dec 5 '18 at 11:05










          • $begingroup$
            @Moshe To prove that we can proceed as follows $$left(frac{3n^2-n+1}{2n^2+n+1}right)^{frac{n^2}{1-n}}=left(frac{3n^2}{2n^2}right)^{-frac{n^2}{n-1}}cdot left(frac{1-1/3n+1/3n^2}{1+1/2n+1/2n^2}right)^{-frac{n^2}{n-1}}$$ and the second part on the RHS tends to 1.
            $endgroup$
            – gimusi
            Dec 5 '18 at 11:16












          • $begingroup$
            Nice and simple (as usual !). Cheers.
            $endgroup$
            – Claude Leibovici
            Dec 5 '18 at 11:25










          • $begingroup$
            @ClaudeLeibovici Thanks Claude! Much appreciative from you! Many cheers :)
            $endgroup$
            – gimusi
            Dec 5 '18 at 11:28










          • $begingroup$
            "[T]he second part on the RHS" does not converge to 1 (hence the equivalence sign in the answer is wrong).
            $endgroup$
            – Did
            Dec 5 '18 at 15:01



















          6












          $begingroup$

          You can estimate:
          $$frac{2n^2+n+1}{3n^2-n+1}<frac{3}{4}, n>7;$$



          $$0<left(frac{3n^2-n+1}{2n^2+n+1}right)^{large frac{n^2}{1-n}}=left(frac{2n^2+n+1}{3n^2-n+1}right)^{large frac{n^2}{n-1}}<left(frac34right)^{large frac{n^2}{n-1}}=left(frac34right)^{n+1+large frac{1}{n-1}}to 0.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            very nice alternative approach (+1).
            $endgroup$
            – gimusi
            Dec 5 '18 at 12:58










          • $begingroup$
            Thank you gimusi.
            $endgroup$
            – farruhota
            Dec 5 '18 at 13:01



















          0












          $begingroup$

          You can't reduce this to the limit form of the exponential function, because the limit of the content of the bracket is 3/2. But the exponent becomes more as $-n$ as $ntoinfty$; so you have the the reciprocal of a positive-exponentially increasing number.






          share|cite|improve this answer









          $endgroup$













            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',
            autoActivateHeartbeat: false,
            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%2f3026929%2flimit-lim-n-to-infty-left-frac3n2-n12n2n1-right-large-fracn%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









            4












            $begingroup$

            HINT



            $$left(frac{3n^2-n+1}{2n^2+n+1}right)^{frac{n^2}{1-n}}sim ccdot left(frac32right)^{-n}$$



            indeed



            $$left(frac{3n^2-n+1}{2n^2+n+1}right)^{frac{n^2}{1-n}}=left(frac32right)^{-n}left(frac{2n^2-frac23n+frac23}{2n^2+n+1}right)^{frac{n^2}{1-n}}$$



            and



            $$left(frac{2n^2-frac23n+frac23}{2n^2+n+1}right)^{frac{n^2}{1-n}}=left[left(1-frac{frac53n+frac13}{2n^2+n+1}right)^{frac{2n^2+n+1}{frac53n+frac13}}right]^{frac{frac53n^3+frac13n^2}{(2n^2+n+1)(1-n)}}toleft(frac1eright)^{-frac56}$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              I observed that, it leads to 0 but subtituing larger values in the original form gives me other value
              $endgroup$
              – Moshe
              Dec 5 '18 at 11:05










            • $begingroup$
              @Moshe To prove that we can proceed as follows $$left(frac{3n^2-n+1}{2n^2+n+1}right)^{frac{n^2}{1-n}}=left(frac{3n^2}{2n^2}right)^{-frac{n^2}{n-1}}cdot left(frac{1-1/3n+1/3n^2}{1+1/2n+1/2n^2}right)^{-frac{n^2}{n-1}}$$ and the second part on the RHS tends to 1.
              $endgroup$
              – gimusi
              Dec 5 '18 at 11:16












            • $begingroup$
              Nice and simple (as usual !). Cheers.
              $endgroup$
              – Claude Leibovici
              Dec 5 '18 at 11:25










            • $begingroup$
              @ClaudeLeibovici Thanks Claude! Much appreciative from you! Many cheers :)
              $endgroup$
              – gimusi
              Dec 5 '18 at 11:28










            • $begingroup$
              "[T]he second part on the RHS" does not converge to 1 (hence the equivalence sign in the answer is wrong).
              $endgroup$
              – Did
              Dec 5 '18 at 15:01
















            4












            $begingroup$

            HINT



            $$left(frac{3n^2-n+1}{2n^2+n+1}right)^{frac{n^2}{1-n}}sim ccdot left(frac32right)^{-n}$$



            indeed



            $$left(frac{3n^2-n+1}{2n^2+n+1}right)^{frac{n^2}{1-n}}=left(frac32right)^{-n}left(frac{2n^2-frac23n+frac23}{2n^2+n+1}right)^{frac{n^2}{1-n}}$$



            and



            $$left(frac{2n^2-frac23n+frac23}{2n^2+n+1}right)^{frac{n^2}{1-n}}=left[left(1-frac{frac53n+frac13}{2n^2+n+1}right)^{frac{2n^2+n+1}{frac53n+frac13}}right]^{frac{frac53n^3+frac13n^2}{(2n^2+n+1)(1-n)}}toleft(frac1eright)^{-frac56}$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              I observed that, it leads to 0 but subtituing larger values in the original form gives me other value
              $endgroup$
              – Moshe
              Dec 5 '18 at 11:05










            • $begingroup$
              @Moshe To prove that we can proceed as follows $$left(frac{3n^2-n+1}{2n^2+n+1}right)^{frac{n^2}{1-n}}=left(frac{3n^2}{2n^2}right)^{-frac{n^2}{n-1}}cdot left(frac{1-1/3n+1/3n^2}{1+1/2n+1/2n^2}right)^{-frac{n^2}{n-1}}$$ and the second part on the RHS tends to 1.
              $endgroup$
              – gimusi
              Dec 5 '18 at 11:16












            • $begingroup$
              Nice and simple (as usual !). Cheers.
              $endgroup$
              – Claude Leibovici
              Dec 5 '18 at 11:25










            • $begingroup$
              @ClaudeLeibovici Thanks Claude! Much appreciative from you! Many cheers :)
              $endgroup$
              – gimusi
              Dec 5 '18 at 11:28










            • $begingroup$
              "[T]he second part on the RHS" does not converge to 1 (hence the equivalence sign in the answer is wrong).
              $endgroup$
              – Did
              Dec 5 '18 at 15:01














            4












            4








            4





            $begingroup$

            HINT



            $$left(frac{3n^2-n+1}{2n^2+n+1}right)^{frac{n^2}{1-n}}sim ccdot left(frac32right)^{-n}$$



            indeed



            $$left(frac{3n^2-n+1}{2n^2+n+1}right)^{frac{n^2}{1-n}}=left(frac32right)^{-n}left(frac{2n^2-frac23n+frac23}{2n^2+n+1}right)^{frac{n^2}{1-n}}$$



            and



            $$left(frac{2n^2-frac23n+frac23}{2n^2+n+1}right)^{frac{n^2}{1-n}}=left[left(1-frac{frac53n+frac13}{2n^2+n+1}right)^{frac{2n^2+n+1}{frac53n+frac13}}right]^{frac{frac53n^3+frac13n^2}{(2n^2+n+1)(1-n)}}toleft(frac1eright)^{-frac56}$$






            share|cite|improve this answer











            $endgroup$



            HINT



            $$left(frac{3n^2-n+1}{2n^2+n+1}right)^{frac{n^2}{1-n}}sim ccdot left(frac32right)^{-n}$$



            indeed



            $$left(frac{3n^2-n+1}{2n^2+n+1}right)^{frac{n^2}{1-n}}=left(frac32right)^{-n}left(frac{2n^2-frac23n+frac23}{2n^2+n+1}right)^{frac{n^2}{1-n}}$$



            and



            $$left(frac{2n^2-frac23n+frac23}{2n^2+n+1}right)^{frac{n^2}{1-n}}=left[left(1-frac{frac53n+frac13}{2n^2+n+1}right)^{frac{2n^2+n+1}{frac53n+frac13}}right]^{frac{frac53n^3+frac13n^2}{(2n^2+n+1)(1-n)}}toleft(frac1eright)^{-frac56}$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 5 '18 at 15:13

























            answered Dec 5 '18 at 10:49









            gimusigimusi

            92.8k84494




            92.8k84494












            • $begingroup$
              I observed that, it leads to 0 but subtituing larger values in the original form gives me other value
              $endgroup$
              – Moshe
              Dec 5 '18 at 11:05










            • $begingroup$
              @Moshe To prove that we can proceed as follows $$left(frac{3n^2-n+1}{2n^2+n+1}right)^{frac{n^2}{1-n}}=left(frac{3n^2}{2n^2}right)^{-frac{n^2}{n-1}}cdot left(frac{1-1/3n+1/3n^2}{1+1/2n+1/2n^2}right)^{-frac{n^2}{n-1}}$$ and the second part on the RHS tends to 1.
              $endgroup$
              – gimusi
              Dec 5 '18 at 11:16












            • $begingroup$
              Nice and simple (as usual !). Cheers.
              $endgroup$
              – Claude Leibovici
              Dec 5 '18 at 11:25










            • $begingroup$
              @ClaudeLeibovici Thanks Claude! Much appreciative from you! Many cheers :)
              $endgroup$
              – gimusi
              Dec 5 '18 at 11:28










            • $begingroup$
              "[T]he second part on the RHS" does not converge to 1 (hence the equivalence sign in the answer is wrong).
              $endgroup$
              – Did
              Dec 5 '18 at 15:01


















            • $begingroup$
              I observed that, it leads to 0 but subtituing larger values in the original form gives me other value
              $endgroup$
              – Moshe
              Dec 5 '18 at 11:05










            • $begingroup$
              @Moshe To prove that we can proceed as follows $$left(frac{3n^2-n+1}{2n^2+n+1}right)^{frac{n^2}{1-n}}=left(frac{3n^2}{2n^2}right)^{-frac{n^2}{n-1}}cdot left(frac{1-1/3n+1/3n^2}{1+1/2n+1/2n^2}right)^{-frac{n^2}{n-1}}$$ and the second part on the RHS tends to 1.
              $endgroup$
              – gimusi
              Dec 5 '18 at 11:16












            • $begingroup$
              Nice and simple (as usual !). Cheers.
              $endgroup$
              – Claude Leibovici
              Dec 5 '18 at 11:25










            • $begingroup$
              @ClaudeLeibovici Thanks Claude! Much appreciative from you! Many cheers :)
              $endgroup$
              – gimusi
              Dec 5 '18 at 11:28










            • $begingroup$
              "[T]he second part on the RHS" does not converge to 1 (hence the equivalence sign in the answer is wrong).
              $endgroup$
              – Did
              Dec 5 '18 at 15:01
















            $begingroup$
            I observed that, it leads to 0 but subtituing larger values in the original form gives me other value
            $endgroup$
            – Moshe
            Dec 5 '18 at 11:05




            $begingroup$
            I observed that, it leads to 0 but subtituing larger values in the original form gives me other value
            $endgroup$
            – Moshe
            Dec 5 '18 at 11:05












            $begingroup$
            @Moshe To prove that we can proceed as follows $$left(frac{3n^2-n+1}{2n^2+n+1}right)^{frac{n^2}{1-n}}=left(frac{3n^2}{2n^2}right)^{-frac{n^2}{n-1}}cdot left(frac{1-1/3n+1/3n^2}{1+1/2n+1/2n^2}right)^{-frac{n^2}{n-1}}$$ and the second part on the RHS tends to 1.
            $endgroup$
            – gimusi
            Dec 5 '18 at 11:16






            $begingroup$
            @Moshe To prove that we can proceed as follows $$left(frac{3n^2-n+1}{2n^2+n+1}right)^{frac{n^2}{1-n}}=left(frac{3n^2}{2n^2}right)^{-frac{n^2}{n-1}}cdot left(frac{1-1/3n+1/3n^2}{1+1/2n+1/2n^2}right)^{-frac{n^2}{n-1}}$$ and the second part on the RHS tends to 1.
            $endgroup$
            – gimusi
            Dec 5 '18 at 11:16














            $begingroup$
            Nice and simple (as usual !). Cheers.
            $endgroup$
            – Claude Leibovici
            Dec 5 '18 at 11:25




            $begingroup$
            Nice and simple (as usual !). Cheers.
            $endgroup$
            – Claude Leibovici
            Dec 5 '18 at 11:25












            $begingroup$
            @ClaudeLeibovici Thanks Claude! Much appreciative from you! Many cheers :)
            $endgroup$
            – gimusi
            Dec 5 '18 at 11:28




            $begingroup$
            @ClaudeLeibovici Thanks Claude! Much appreciative from you! Many cheers :)
            $endgroup$
            – gimusi
            Dec 5 '18 at 11:28












            $begingroup$
            "[T]he second part on the RHS" does not converge to 1 (hence the equivalence sign in the answer is wrong).
            $endgroup$
            – Did
            Dec 5 '18 at 15:01




            $begingroup$
            "[T]he second part on the RHS" does not converge to 1 (hence the equivalence sign in the answer is wrong).
            $endgroup$
            – Did
            Dec 5 '18 at 15:01











            6












            $begingroup$

            You can estimate:
            $$frac{2n^2+n+1}{3n^2-n+1}<frac{3}{4}, n>7;$$



            $$0<left(frac{3n^2-n+1}{2n^2+n+1}right)^{large frac{n^2}{1-n}}=left(frac{2n^2+n+1}{3n^2-n+1}right)^{large frac{n^2}{n-1}}<left(frac34right)^{large frac{n^2}{n-1}}=left(frac34right)^{n+1+large frac{1}{n-1}}to 0.$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              very nice alternative approach (+1).
              $endgroup$
              – gimusi
              Dec 5 '18 at 12:58










            • $begingroup$
              Thank you gimusi.
              $endgroup$
              – farruhota
              Dec 5 '18 at 13:01
















            6












            $begingroup$

            You can estimate:
            $$frac{2n^2+n+1}{3n^2-n+1}<frac{3}{4}, n>7;$$



            $$0<left(frac{3n^2-n+1}{2n^2+n+1}right)^{large frac{n^2}{1-n}}=left(frac{2n^2+n+1}{3n^2-n+1}right)^{large frac{n^2}{n-1}}<left(frac34right)^{large frac{n^2}{n-1}}=left(frac34right)^{n+1+large frac{1}{n-1}}to 0.$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              very nice alternative approach (+1).
              $endgroup$
              – gimusi
              Dec 5 '18 at 12:58










            • $begingroup$
              Thank you gimusi.
              $endgroup$
              – farruhota
              Dec 5 '18 at 13:01














            6












            6








            6





            $begingroup$

            You can estimate:
            $$frac{2n^2+n+1}{3n^2-n+1}<frac{3}{4}, n>7;$$



            $$0<left(frac{3n^2-n+1}{2n^2+n+1}right)^{large frac{n^2}{1-n}}=left(frac{2n^2+n+1}{3n^2-n+1}right)^{large frac{n^2}{n-1}}<left(frac34right)^{large frac{n^2}{n-1}}=left(frac34right)^{n+1+large frac{1}{n-1}}to 0.$$






            share|cite|improve this answer









            $endgroup$



            You can estimate:
            $$frac{2n^2+n+1}{3n^2-n+1}<frac{3}{4}, n>7;$$



            $$0<left(frac{3n^2-n+1}{2n^2+n+1}right)^{large frac{n^2}{1-n}}=left(frac{2n^2+n+1}{3n^2-n+1}right)^{large frac{n^2}{n-1}}<left(frac34right)^{large frac{n^2}{n-1}}=left(frac34right)^{n+1+large frac{1}{n-1}}to 0.$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 5 '18 at 12:15









            farruhotafarruhota

            20k2738




            20k2738












            • $begingroup$
              very nice alternative approach (+1).
              $endgroup$
              – gimusi
              Dec 5 '18 at 12:58










            • $begingroup$
              Thank you gimusi.
              $endgroup$
              – farruhota
              Dec 5 '18 at 13:01


















            • $begingroup$
              very nice alternative approach (+1).
              $endgroup$
              – gimusi
              Dec 5 '18 at 12:58










            • $begingroup$
              Thank you gimusi.
              $endgroup$
              – farruhota
              Dec 5 '18 at 13:01
















            $begingroup$
            very nice alternative approach (+1).
            $endgroup$
            – gimusi
            Dec 5 '18 at 12:58




            $begingroup$
            very nice alternative approach (+1).
            $endgroup$
            – gimusi
            Dec 5 '18 at 12:58












            $begingroup$
            Thank you gimusi.
            $endgroup$
            – farruhota
            Dec 5 '18 at 13:01




            $begingroup$
            Thank you gimusi.
            $endgroup$
            – farruhota
            Dec 5 '18 at 13:01











            0












            $begingroup$

            You can't reduce this to the limit form of the exponential function, because the limit of the content of the bracket is 3/2. But the exponent becomes more as $-n$ as $ntoinfty$; so you have the the reciprocal of a positive-exponentially increasing number.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              You can't reduce this to the limit form of the exponential function, because the limit of the content of the bracket is 3/2. But the exponent becomes more as $-n$ as $ntoinfty$; so you have the the reciprocal of a positive-exponentially increasing number.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                You can't reduce this to the limit form of the exponential function, because the limit of the content of the bracket is 3/2. But the exponent becomes more as $-n$ as $ntoinfty$; so you have the the reciprocal of a positive-exponentially increasing number.






                share|cite|improve this answer









                $endgroup$



                You can't reduce this to the limit form of the exponential function, because the limit of the content of the bracket is 3/2. But the exponent becomes more as $-n$ as $ntoinfty$; so you have the the reciprocal of a positive-exponentially increasing number.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 5 '18 at 14:14









                AmbretteOrriseyAmbretteOrrisey

                54210




                54210






























                    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.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3026929%2flimit-lim-n-to-infty-left-frac3n2-n12n2n1-right-large-fracn%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

                    Bundesstraße 106

                    Image
                    Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück
                    Read more

                    Verónica Boquete

                    Image
                    Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl
                    Read more

                    Ida-Boy-Ed-Garten

                    Image
                    Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
                    Read more