Calculate the limit $lim_{ntoinfty } sqrt[n]{4^{n}-3^{n-2}} $












1














I am trying to calculate the limit:
$$lim_{ntoinfty}sqrt[n]{4^{n}-3^{n-2}}$$ My progression so far is rewiring as:



$$lim_{ntoinfty}left(4^{n}-3^{n}cdotfrac{1}{9}right)^{frac{1}{n}}$$



Are my steps correct? How do I continue?






limits







share|cite|improve this question





























    1














    I am trying to calculate the limit:
    $$lim_{ntoinfty}sqrt[n]{4^{n}-3^{n-2}}$$ My progression so far is rewiring as:



    $$lim_{ntoinfty}left(4^{n}-3^{n}cdotfrac{1}{9}right)^{frac{1}{n}}$$



    Are my steps correct? How do I continue?






    limits







    share|cite|improve this question



























      1












      1








      1


      0





      I am trying to calculate the limit:
      $$lim_{ntoinfty}sqrt[n]{4^{n}-3^{n-2}}$$ My progression so far is rewiring as:



      $$lim_{ntoinfty}left(4^{n}-3^{n}cdotfrac{1}{9}right)^{frac{1}{n}}$$



      Are my steps correct? How do I continue?






      limits







      share|cite|improve this question















      I am trying to calculate the limit:
      $$lim_{ntoinfty}sqrt[n]{4^{n}-3^{n-2}}$$ My progression so far is rewiring as:



      $$lim_{ntoinfty}left(4^{n}-3^{n}cdotfrac{1}{9}right)^{frac{1}{n}}$$



      Are my steps correct? How do I continue?






      limits




      limits






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 23 at 10:01









      Mutantoe

      564411




      564411










      asked Nov 20 at 18:16









      Jack

      213




      213






















          3 Answers
          3






          active

          oldest

          votes


















          2














          Hint: Rewrite as $$lim_{n to infty} sqrt[n]{4^n - 3^{n-2}} = lim_{n to infty} 4 sqrt[n] {1 - frac{1}{9} left( frac{3}{4}right)^n}$$ and consider the limiting behavior of the term under the radical sign.






          share|cite|improve this answer

















          • 1




            so the answer is 4?
            – Jack
            Nov 20 at 18:35










          • Yes, it is. If $f(n) to 1$, then $sqrt[n]{f(n)} to 1$ as well.
            – Connor Harris
            Nov 20 at 18:39





















          1














          HINT



          We have that



          $$sqrt[n]{4^{n}-3^{n-2}}sim sqrt[n]{4^{n}}$$






          share|cite|improve this answer





















          • so the answer is 4?
            – Jack
            Nov 20 at 18:35










          • Yes of course indeed $4^n>>3^{n-2}$ that is $$frac{4^n-3^{n-2}}{4^n}to 1$$
            – gimusi
            Nov 20 at 18:36



















          0














          You can also evaluate the limit in the following way



          $$begin{aligned}lim_{ntoinfty}sqrt[n]{4^n-3^{n-2}} &= lim_{ntoinfty}4left({1-frac{1}{9}left( frac{3}{4}right)^n}right)^frac1n=4lim_{ntoinfty}left(1-frac1{9left(frac43right)^n}right)^frac1n=\
          &=4lim_{ntoinfty}left(1-frac1{9left(frac43right)^n}right)^{9left(frac43right)^nfrac19left(frac34right)^nfrac1n}=\
          &=4left(lim_{ntoinfty}left(1-frac1{9left(frac43right)^n}right)^{9left(frac43right)^n}right)^{lim_{ntoinfty}frac19left(frac34right)^nfrac1n}=\
          &=4left(e^{-1}right)^0=4
          end{aligned}$$






          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',
            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%2f3006693%2fcalculate-the-limit-lim-n-to-infty-sqrtn4n-3n-2%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









            2














            Hint: Rewrite as $$lim_{n to infty} sqrt[n]{4^n - 3^{n-2}} = lim_{n to infty} 4 sqrt[n] {1 - frac{1}{9} left( frac{3}{4}right)^n}$$ and consider the limiting behavior of the term under the radical sign.






            share|cite|improve this answer

















            • 1




              so the answer is 4?
              – Jack
              Nov 20 at 18:35










            • Yes, it is. If $f(n) to 1$, then $sqrt[n]{f(n)} to 1$ as well.
              – Connor Harris
              Nov 20 at 18:39


















            2














            Hint: Rewrite as $$lim_{n to infty} sqrt[n]{4^n - 3^{n-2}} = lim_{n to infty} 4 sqrt[n] {1 - frac{1}{9} left( frac{3}{4}right)^n}$$ and consider the limiting behavior of the term under the radical sign.






            share|cite|improve this answer

















            • 1




              so the answer is 4?
              – Jack
              Nov 20 at 18:35










            • Yes, it is. If $f(n) to 1$, then $sqrt[n]{f(n)} to 1$ as well.
              – Connor Harris
              Nov 20 at 18:39
















            2












            2








            2






            Hint: Rewrite as $$lim_{n to infty} sqrt[n]{4^n - 3^{n-2}} = lim_{n to infty} 4 sqrt[n] {1 - frac{1}{9} left( frac{3}{4}right)^n}$$ and consider the limiting behavior of the term under the radical sign.






            share|cite|improve this answer












            Hint: Rewrite as $$lim_{n to infty} sqrt[n]{4^n - 3^{n-2}} = lim_{n to infty} 4 sqrt[n] {1 - frac{1}{9} left( frac{3}{4}right)^n}$$ and consider the limiting behavior of the term under the radical sign.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 20 at 18:19









            Connor Harris

            4,303723




            4,303723








            • 1




              so the answer is 4?
              – Jack
              Nov 20 at 18:35










            • Yes, it is. If $f(n) to 1$, then $sqrt[n]{f(n)} to 1$ as well.
              – Connor Harris
              Nov 20 at 18:39
















            • 1




              so the answer is 4?
              – Jack
              Nov 20 at 18:35










            • Yes, it is. If $f(n) to 1$, then $sqrt[n]{f(n)} to 1$ as well.
              – Connor Harris
              Nov 20 at 18:39










            1




            1




            so the answer is 4?
            – Jack
            Nov 20 at 18:35




            so the answer is 4?
            – Jack
            Nov 20 at 18:35












            Yes, it is. If $f(n) to 1$, then $sqrt[n]{f(n)} to 1$ as well.
            – Connor Harris
            Nov 20 at 18:39






            Yes, it is. If $f(n) to 1$, then $sqrt[n]{f(n)} to 1$ as well.
            – Connor Harris
            Nov 20 at 18:39













            1














            HINT



            We have that



            $$sqrt[n]{4^{n}-3^{n-2}}sim sqrt[n]{4^{n}}$$






            share|cite|improve this answer





















            • so the answer is 4?
              – Jack
              Nov 20 at 18:35










            • Yes of course indeed $4^n>>3^{n-2}$ that is $$frac{4^n-3^{n-2}}{4^n}to 1$$
              – gimusi
              Nov 20 at 18:36
















            1














            HINT



            We have that



            $$sqrt[n]{4^{n}-3^{n-2}}sim sqrt[n]{4^{n}}$$






            share|cite|improve this answer





















            • so the answer is 4?
              – Jack
              Nov 20 at 18:35










            • Yes of course indeed $4^n>>3^{n-2}$ that is $$frac{4^n-3^{n-2}}{4^n}to 1$$
              – gimusi
              Nov 20 at 18:36














            1












            1








            1






            HINT



            We have that



            $$sqrt[n]{4^{n}-3^{n-2}}sim sqrt[n]{4^{n}}$$






            share|cite|improve this answer












            HINT



            We have that



            $$sqrt[n]{4^{n}-3^{n-2}}sim sqrt[n]{4^{n}}$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 20 at 18:21









            gimusi

            1




            1












            • so the answer is 4?
              – Jack
              Nov 20 at 18:35










            • Yes of course indeed $4^n>>3^{n-2}$ that is $$frac{4^n-3^{n-2}}{4^n}to 1$$
              – gimusi
              Nov 20 at 18:36


















            • so the answer is 4?
              – Jack
              Nov 20 at 18:35










            • Yes of course indeed $4^n>>3^{n-2}$ that is $$frac{4^n-3^{n-2}}{4^n}to 1$$
              – gimusi
              Nov 20 at 18:36
















            so the answer is 4?
            – Jack
            Nov 20 at 18:35




            so the answer is 4?
            – Jack
            Nov 20 at 18:35












            Yes of course indeed $4^n>>3^{n-2}$ that is $$frac{4^n-3^{n-2}}{4^n}to 1$$
            – gimusi
            Nov 20 at 18:36




            Yes of course indeed $4^n>>3^{n-2}$ that is $$frac{4^n-3^{n-2}}{4^n}to 1$$
            – gimusi
            Nov 20 at 18:36











            0














            You can also evaluate the limit in the following way



            $$begin{aligned}lim_{ntoinfty}sqrt[n]{4^n-3^{n-2}} &= lim_{ntoinfty}4left({1-frac{1}{9}left( frac{3}{4}right)^n}right)^frac1n=4lim_{ntoinfty}left(1-frac1{9left(frac43right)^n}right)^frac1n=\
            &=4lim_{ntoinfty}left(1-frac1{9left(frac43right)^n}right)^{9left(frac43right)^nfrac19left(frac34right)^nfrac1n}=\
            &=4left(lim_{ntoinfty}left(1-frac1{9left(frac43right)^n}right)^{9left(frac43right)^n}right)^{lim_{ntoinfty}frac19left(frac34right)^nfrac1n}=\
            &=4left(e^{-1}right)^0=4
            end{aligned}$$






            share|cite|improve this answer


























              0














              You can also evaluate the limit in the following way



              $$begin{aligned}lim_{ntoinfty}sqrt[n]{4^n-3^{n-2}} &= lim_{ntoinfty}4left({1-frac{1}{9}left( frac{3}{4}right)^n}right)^frac1n=4lim_{ntoinfty}left(1-frac1{9left(frac43right)^n}right)^frac1n=\
              &=4lim_{ntoinfty}left(1-frac1{9left(frac43right)^n}right)^{9left(frac43right)^nfrac19left(frac34right)^nfrac1n}=\
              &=4left(lim_{ntoinfty}left(1-frac1{9left(frac43right)^n}right)^{9left(frac43right)^n}right)^{lim_{ntoinfty}frac19left(frac34right)^nfrac1n}=\
              &=4left(e^{-1}right)^0=4
              end{aligned}$$






              share|cite|improve this answer
























                0












                0








                0






                You can also evaluate the limit in the following way



                $$begin{aligned}lim_{ntoinfty}sqrt[n]{4^n-3^{n-2}} &= lim_{ntoinfty}4left({1-frac{1}{9}left( frac{3}{4}right)^n}right)^frac1n=4lim_{ntoinfty}left(1-frac1{9left(frac43right)^n}right)^frac1n=\
                &=4lim_{ntoinfty}left(1-frac1{9left(frac43right)^n}right)^{9left(frac43right)^nfrac19left(frac34right)^nfrac1n}=\
                &=4left(lim_{ntoinfty}left(1-frac1{9left(frac43right)^n}right)^{9left(frac43right)^n}right)^{lim_{ntoinfty}frac19left(frac34right)^nfrac1n}=\
                &=4left(e^{-1}right)^0=4
                end{aligned}$$






                share|cite|improve this answer












                You can also evaluate the limit in the following way



                $$begin{aligned}lim_{ntoinfty}sqrt[n]{4^n-3^{n-2}} &= lim_{ntoinfty}4left({1-frac{1}{9}left( frac{3}{4}right)^n}right)^frac1n=4lim_{ntoinfty}left(1-frac1{9left(frac43right)^n}right)^frac1n=\
                &=4lim_{ntoinfty}left(1-frac1{9left(frac43right)^n}right)^{9left(frac43right)^nfrac19left(frac34right)^nfrac1n}=\
                &=4left(lim_{ntoinfty}left(1-frac1{9left(frac43right)^n}right)^{9left(frac43right)^n}right)^{lim_{ntoinfty}frac19left(frac34right)^nfrac1n}=\
                &=4left(e^{-1}right)^0=4
                end{aligned}$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 20 at 18:46









                Mikalai Parshutsich

                453315




                453315






























                    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%2f3006693%2fcalculate-the-limit-lim-n-to-infty-sqrtn4n-3n-2%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

                    How to change which sound is reproduced for terminal bell?

                    Image
                    3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question
                    Read more

                    Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

                    Image
                    up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx
                    Read more

                    Can I use Tabulator js library in my java Spring + Thymeleaf project?

                    Image
                    0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"
                    Read more