How to make this limit question a indeterminate form? (L-Hopital)











up vote
3
down vote

favorite












$$lim_{xto 1^+}[ln(x^7 -1) - ln(x^5 -1)]$$



This is a question from L-Hospital rule question set. My approach was to apply log property in this question and solve it, but $ln(frac{0}{0})$ might not be right way to convert it into indeterminate form.



How to solve this question?










share|cite|improve this question









New contributor




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
























    up vote
    3
    down vote

    favorite












    $$lim_{xto 1^+}[ln(x^7 -1) - ln(x^5 -1)]$$



    This is a question from L-Hospital rule question set. My approach was to apply log property in this question and solve it, but $ln(frac{0}{0})$ might not be right way to convert it into indeterminate form.



    How to solve this question?










    share|cite|improve this question









    New contributor




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






















      up vote
      3
      down vote

      favorite









      up vote
      3
      down vote

      favorite











      $$lim_{xto 1^+}[ln(x^7 -1) - ln(x^5 -1)]$$



      This is a question from L-Hospital rule question set. My approach was to apply log property in this question and solve it, but $ln(frac{0}{0})$ might not be right way to convert it into indeterminate form.



      How to solve this question?










      share|cite|improve this question









      New contributor




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











      $$lim_{xto 1^+}[ln(x^7 -1) - ln(x^5 -1)]$$



      This is a question from L-Hospital rule question set. My approach was to apply log property in this question and solve it, but $ln(frac{0}{0})$ might not be right way to convert it into indeterminate form.



      How to solve this question?







      limits






      share|cite|improve this question









      New contributor




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











      share|cite|improve this question









      New contributor




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









      share|cite|improve this question




      share|cite|improve this question








      edited 18 hours ago









      Ng Chung Tak

      13.5k31234




      13.5k31234






      New contributor




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









      asked 19 hours ago









      Amogh Joshi

      183




      183




      New contributor




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





      New contributor





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






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






















          2 Answers
          2






          active

          oldest

          votes

















          up vote
          3
          down vote



          accepted










          $$lim_{x to 1^+} [ln(x^7-1)- ln(x^5-1)]=ln left[lim_{xto 1^+} frac{x^7-1}{x^5-1} right]$$



          Evaluate $lim_{xto 1^+} frac{x^7-1}{x^5-1} $ first and I think you can solve the problem.






          share|cite|improve this answer





















          • I am getting the answer ln(7/5). Is it right?
            – Amogh Joshi
            18 hours ago










          • yes, congratulations.
            – Siong Thye Goh
            18 hours ago


















          up vote
          4
          down vote













          Hint:



          $$lim_{xto 1^+}left[ln(x^7 -1) - ln(x^5 -1)right]=lnleft(lim_{xto 1^+}dfrac{x^7-1}{x^5-1}right)$$



          Now use $x^n-1=(x-1)(x^{n-1}+x^{n-2}+cdots+x+1)$



          Alternatively, $dfrac{x^7-1}{x^5-1}=dfrac{dfrac{x^7-1}{x-1}}{dfrac{x^5-1}{x-1}}$



          and $$lim_{xto 1}dfrac{x^n-1}{x-1}=dfrac{d(x^n)}{dx}_{text{( at }x=1)}$$






          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
            });


            }
            });






            Amogh Joshi is a new contributor. Be nice, and check out our Code of Conduct.










             

            draft saved


            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2997880%2fhow-to-make-this-limit-question-a-indeterminate-form-l-hopital%23new-answer', 'question_page');
            }
            );

            Post as a guest
































            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            3
            down vote



            accepted










            $$lim_{x to 1^+} [ln(x^7-1)- ln(x^5-1)]=ln left[lim_{xto 1^+} frac{x^7-1}{x^5-1} right]$$



            Evaluate $lim_{xto 1^+} frac{x^7-1}{x^5-1} $ first and I think you can solve the problem.






            share|cite|improve this answer





















            • I am getting the answer ln(7/5). Is it right?
              – Amogh Joshi
              18 hours ago










            • yes, congratulations.
              – Siong Thye Goh
              18 hours ago















            up vote
            3
            down vote



            accepted










            $$lim_{x to 1^+} [ln(x^7-1)- ln(x^5-1)]=ln left[lim_{xto 1^+} frac{x^7-1}{x^5-1} right]$$



            Evaluate $lim_{xto 1^+} frac{x^7-1}{x^5-1} $ first and I think you can solve the problem.






            share|cite|improve this answer





















            • I am getting the answer ln(7/5). Is it right?
              – Amogh Joshi
              18 hours ago










            • yes, congratulations.
              – Siong Thye Goh
              18 hours ago













            up vote
            3
            down vote



            accepted







            up vote
            3
            down vote



            accepted






            $$lim_{x to 1^+} [ln(x^7-1)- ln(x^5-1)]=ln left[lim_{xto 1^+} frac{x^7-1}{x^5-1} right]$$



            Evaluate $lim_{xto 1^+} frac{x^7-1}{x^5-1} $ first and I think you can solve the problem.






            share|cite|improve this answer












            $$lim_{x to 1^+} [ln(x^7-1)- ln(x^5-1)]=ln left[lim_{xto 1^+} frac{x^7-1}{x^5-1} right]$$



            Evaluate $lim_{xto 1^+} frac{x^7-1}{x^5-1} $ first and I think you can solve the problem.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 18 hours ago









            Siong Thye Goh

            92.3k1461114




            92.3k1461114












            • I am getting the answer ln(7/5). Is it right?
              – Amogh Joshi
              18 hours ago










            • yes, congratulations.
              – Siong Thye Goh
              18 hours ago


















            • I am getting the answer ln(7/5). Is it right?
              – Amogh Joshi
              18 hours ago










            • yes, congratulations.
              – Siong Thye Goh
              18 hours ago
















            I am getting the answer ln(7/5). Is it right?
            – Amogh Joshi
            18 hours ago




            I am getting the answer ln(7/5). Is it right?
            – Amogh Joshi
            18 hours ago












            yes, congratulations.
            – Siong Thye Goh
            18 hours ago




            yes, congratulations.
            – Siong Thye Goh
            18 hours ago










            up vote
            4
            down vote













            Hint:



            $$lim_{xto 1^+}left[ln(x^7 -1) - ln(x^5 -1)right]=lnleft(lim_{xto 1^+}dfrac{x^7-1}{x^5-1}right)$$



            Now use $x^n-1=(x-1)(x^{n-1}+x^{n-2}+cdots+x+1)$



            Alternatively, $dfrac{x^7-1}{x^5-1}=dfrac{dfrac{x^7-1}{x-1}}{dfrac{x^5-1}{x-1}}$



            and $$lim_{xto 1}dfrac{x^n-1}{x-1}=dfrac{d(x^n)}{dx}_{text{( at }x=1)}$$






            share|cite|improve this answer



























              up vote
              4
              down vote













              Hint:



              $$lim_{xto 1^+}left[ln(x^7 -1) - ln(x^5 -1)right]=lnleft(lim_{xto 1^+}dfrac{x^7-1}{x^5-1}right)$$



              Now use $x^n-1=(x-1)(x^{n-1}+x^{n-2}+cdots+x+1)$



              Alternatively, $dfrac{x^7-1}{x^5-1}=dfrac{dfrac{x^7-1}{x-1}}{dfrac{x^5-1}{x-1}}$



              and $$lim_{xto 1}dfrac{x^n-1}{x-1}=dfrac{d(x^n)}{dx}_{text{( at }x=1)}$$






              share|cite|improve this answer

























                up vote
                4
                down vote










                up vote
                4
                down vote









                Hint:



                $$lim_{xto 1^+}left[ln(x^7 -1) - ln(x^5 -1)right]=lnleft(lim_{xto 1^+}dfrac{x^7-1}{x^5-1}right)$$



                Now use $x^n-1=(x-1)(x^{n-1}+x^{n-2}+cdots+x+1)$



                Alternatively, $dfrac{x^7-1}{x^5-1}=dfrac{dfrac{x^7-1}{x-1}}{dfrac{x^5-1}{x-1}}$



                and $$lim_{xto 1}dfrac{x^n-1}{x-1}=dfrac{d(x^n)}{dx}_{text{( at }x=1)}$$






                share|cite|improve this answer














                Hint:



                $$lim_{xto 1^+}left[ln(x^7 -1) - ln(x^5 -1)right]=lnleft(lim_{xto 1^+}dfrac{x^7-1}{x^5-1}right)$$



                Now use $x^n-1=(x-1)(x^{n-1}+x^{n-2}+cdots+x+1)$



                Alternatively, $dfrac{x^7-1}{x^5-1}=dfrac{dfrac{x^7-1}{x-1}}{dfrac{x^5-1}{x-1}}$



                and $$lim_{xto 1}dfrac{x^n-1}{x-1}=dfrac{d(x^n)}{dx}_{text{( at }x=1)}$$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited 18 hours ago

























                answered 18 hours ago









                lab bhattacharjee

                219k14154270




                219k14154270






















                    Amogh Joshi is a new contributor. Be nice, and check out our Code of Conduct.










                     

                    draft saved


                    draft discarded


















                    Amogh Joshi is a new contributor. Be nice, and check out our Code of Conduct.













                    Amogh Joshi is a new contributor. Be nice, and check out our Code of Conduct.












                    Amogh Joshi is a new contributor. Be nice, and check out our Code of Conduct.















                     


                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2997880%2fhow-to-make-this-limit-question-a-indeterminate-form-l-hopital%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest




















































































                    Popular posts from this blog

                    Plaza Victoria

                    Puebla de Zaragoza

                    Musa