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











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



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







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    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?






    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.






















      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?






      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.











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



      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




      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






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      Amogh Joshi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      asked 19 hours ago









      Amogh Joshi

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      New contributor




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      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.






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



          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























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            2 Answers
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            active

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            2 Answers
            2






            active

            oldest

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            active

            oldest

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            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






















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