Limit Question: doubting about my answer! A limit where $x to infty$












2












$begingroup$


Hey guys so I have this limit:



$$lim_{x to ∞} f(x) = frac{(3x+1)^3(x-1)}{(x-2)^4}$$



and I got $27$ as final answer, just wondering if you guys can check;
I expanded the numerator and denominator and then divided everything by $x^4$






limits







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    Hey guys so I have this limit:



    $$lim_{x to ∞} f(x) = frac{(3x+1)^3(x-1)}{(x-2)^4}$$



    and I got $27$ as final answer, just wondering if you guys can check;
    I expanded the numerator and denominator and then divided everything by $x^4$






    limits







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      Hey guys so I have this limit:



      $$lim_{x to ∞} f(x) = frac{(3x+1)^3(x-1)}{(x-2)^4}$$



      and I got $27$ as final answer, just wondering if you guys can check;
      I expanded the numerator and denominator and then divided everything by $x^4$






      limits







      share|cite|improve this question











      $endgroup$




      Hey guys so I have this limit:



      $$lim_{x to ∞} f(x) = frac{(3x+1)^3(x-1)}{(x-2)^4}$$



      and I got $27$ as final answer, just wondering if you guys can check;
      I expanded the numerator and denominator and then divided everything by $x^4$






      limits




      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 5:20









      Eevee Trainer

      7,02311337




      7,02311337










      asked Dec 5 '18 at 5:08









      S..S..

      545




      545






















          6 Answers
          6






          active

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          4












          $begingroup$

          Rather than expanding, divide top and bottom by $x^4$ directly.



          $$lim_{x to infty}frac{(3x+1)^3(x-1)}{(x-2)^4}=lim_{x to infty}frac{(3+frac1x)^3(1-frac1x)}{(1-frac2x)^4}=frac{27(1)}1=27$$



          $27$ is the correct answer.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            Waaat, need to practice that; I waste to much time expanding things
            $endgroup$
            – S..
            Dec 5 '18 at 5:17



















          3












          $begingroup$

          Note that



          $$frac{(3x+1)^3(x-1)}{(x-2)^4}=27frac{(x+1/3)^3(x-1)}{(x-2)^4}=27frac{frac{(x+1/3)^3}{x^3}frac{(x-1)}x}{frac{(x-2)^4}{x^4}}\=27frac{(1+1/x)^3(1-1/x)}{(1-2/x)^4}to27frac{(1+0)^3(1-0)}{(1-0)^4}=27$$



          as $xtoinfty$, so your result is correct.






          share|cite|improve this answer









          $endgroup$





















            3












            $begingroup$

            Yes, your answer is correct. An easy way is



            begin{align}
            lim_{x to ∞} f(x) &=lim_{x to ∞} frac{(3x+1)^3(x-1)}{(x-2)^4}\
            &= lim_{x to ∞}frac{x^4cdot (3+frac{1}{x})^3(1-frac{1}{x})}{x^4cdot (1-frac{2}{x})^4}\
            &= lim_{x to ∞} frac{(3+frac{1}{x})^3(1-frac{1}{x})}{ (1-frac{2}{x})^4}\
            &=27
            end{align}






            share|cite|improve this answer









            $endgroup$





















              2












              $begingroup$

              You don't need to really check too much. It's sufficient to see that:



              $$(3x+1)^3(x-1) = 3^3x^4 + text{(other stuff of a lesser degree)}$$



              and



              $$(x-2)^4 = x^4 + text{(other stuff of a lesser degree)}$$



              Since the numerator and denominator have the same degree, the limit as $xto infty$ will be the ratio of the highest-degree terms' coefficients. Since the coefficients of the $x^4$ terms, respectively, are $3^3 = 27$ and $1$, then the limit evaluates to



              $$lim_{x to ∞} f(x) = frac{(3x+1)^3(x-1)}{(x-2)^4} = frac{27}{1} = 27$$



              So, tl;dr version, you're right. The argument could be made that you did a bit more work than necessary by multiplying everything out, but that's more an issue of personal taste in a case like this (since it's not a ton more work).






              share|cite|improve this answer









              $endgroup$





















                1












                $begingroup$

                Only to check or guess the possible result we can use that the constant term are negligeble then



                $$frac{(3x+1)^3(x-1)}{(x-2)^4}sim frac{(3x)^3(x)}{(x)^4}=frac{27x^4}{x^4}=27$$



                and then proceed for a rigorous proof as already indicated.






                share|cite|improve this answer









                $endgroup$





















                  1












                  $begingroup$

                  As an alternative method (for your practicing) you can rearrange degrees:
                  $$lim_{x to ∞} frac{(3x+1)^3(x-1)}{(x-2)^4}=lim_{x to ∞} left(frac{3x+1}{x-2}right)^3left(frac{x-1}{x-2}right)=cdots=3^3cdot 1=27.$$
                  Note: In the dots you should use the properies of limits.






                  share|cite|improve this answer









                  $endgroup$













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                    6 Answers
                    6






                    active

                    oldest

                    votes








                    6 Answers
                    6






                    active

                    oldest

                    votes









                    active

                    oldest

                    votes






                    active

                    oldest

                    votes









                    4












                    $begingroup$

                    Rather than expanding, divide top and bottom by $x^4$ directly.



                    $$lim_{x to infty}frac{(3x+1)^3(x-1)}{(x-2)^4}=lim_{x to infty}frac{(3+frac1x)^3(1-frac1x)}{(1-frac2x)^4}=frac{27(1)}1=27$$



                    $27$ is the correct answer.






                    share|cite|improve this answer









                    $endgroup$









                    • 1




                      $begingroup$
                      Waaat, need to practice that; I waste to much time expanding things
                      $endgroup$
                      – S..
                      Dec 5 '18 at 5:17
















                    4












                    $begingroup$

                    Rather than expanding, divide top and bottom by $x^4$ directly.



                    $$lim_{x to infty}frac{(3x+1)^3(x-1)}{(x-2)^4}=lim_{x to infty}frac{(3+frac1x)^3(1-frac1x)}{(1-frac2x)^4}=frac{27(1)}1=27$$



                    $27$ is the correct answer.






                    share|cite|improve this answer









                    $endgroup$









                    • 1




                      $begingroup$
                      Waaat, need to practice that; I waste to much time expanding things
                      $endgroup$
                      – S..
                      Dec 5 '18 at 5:17














                    4












                    4








                    4





                    $begingroup$

                    Rather than expanding, divide top and bottom by $x^4$ directly.



                    $$lim_{x to infty}frac{(3x+1)^3(x-1)}{(x-2)^4}=lim_{x to infty}frac{(3+frac1x)^3(1-frac1x)}{(1-frac2x)^4}=frac{27(1)}1=27$$



                    $27$ is the correct answer.






                    share|cite|improve this answer









                    $endgroup$



                    Rather than expanding, divide top and bottom by $x^4$ directly.



                    $$lim_{x to infty}frac{(3x+1)^3(x-1)}{(x-2)^4}=lim_{x to infty}frac{(3+frac1x)^3(1-frac1x)}{(1-frac2x)^4}=frac{27(1)}1=27$$



                    $27$ is the correct answer.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 5 '18 at 5:12









                    Siong Thye GohSiong Thye Goh

                    102k1466118




                    102k1466118








                    • 1




                      $begingroup$
                      Waaat, need to practice that; I waste to much time expanding things
                      $endgroup$
                      – S..
                      Dec 5 '18 at 5:17














                    • 1




                      $begingroup$
                      Waaat, need to practice that; I waste to much time expanding things
                      $endgroup$
                      – S..
                      Dec 5 '18 at 5:17








                    1




                    1




                    $begingroup$
                    Waaat, need to practice that; I waste to much time expanding things
                    $endgroup$
                    – S..
                    Dec 5 '18 at 5:17




                    $begingroup$
                    Waaat, need to practice that; I waste to much time expanding things
                    $endgroup$
                    – S..
                    Dec 5 '18 at 5:17











                    3












                    $begingroup$

                    Note that



                    $$frac{(3x+1)^3(x-1)}{(x-2)^4}=27frac{(x+1/3)^3(x-1)}{(x-2)^4}=27frac{frac{(x+1/3)^3}{x^3}frac{(x-1)}x}{frac{(x-2)^4}{x^4}}\=27frac{(1+1/x)^3(1-1/x)}{(1-2/x)^4}to27frac{(1+0)^3(1-0)}{(1-0)^4}=27$$



                    as $xtoinfty$, so your result is correct.






                    share|cite|improve this answer









                    $endgroup$


















                      3












                      $begingroup$

                      Note that



                      $$frac{(3x+1)^3(x-1)}{(x-2)^4}=27frac{(x+1/3)^3(x-1)}{(x-2)^4}=27frac{frac{(x+1/3)^3}{x^3}frac{(x-1)}x}{frac{(x-2)^4}{x^4}}\=27frac{(1+1/x)^3(1-1/x)}{(1-2/x)^4}to27frac{(1+0)^3(1-0)}{(1-0)^4}=27$$



                      as $xtoinfty$, so your result is correct.






                      share|cite|improve this answer









                      $endgroup$
















                        3












                        3








                        3





                        $begingroup$

                        Note that



                        $$frac{(3x+1)^3(x-1)}{(x-2)^4}=27frac{(x+1/3)^3(x-1)}{(x-2)^4}=27frac{frac{(x+1/3)^3}{x^3}frac{(x-1)}x}{frac{(x-2)^4}{x^4}}\=27frac{(1+1/x)^3(1-1/x)}{(1-2/x)^4}to27frac{(1+0)^3(1-0)}{(1-0)^4}=27$$



                        as $xtoinfty$, so your result is correct.






                        share|cite|improve this answer









                        $endgroup$



                        Note that



                        $$frac{(3x+1)^3(x-1)}{(x-2)^4}=27frac{(x+1/3)^3(x-1)}{(x-2)^4}=27frac{frac{(x+1/3)^3}{x^3}frac{(x-1)}x}{frac{(x-2)^4}{x^4}}\=27frac{(1+1/x)^3(1-1/x)}{(1-2/x)^4}to27frac{(1+0)^3(1-0)}{(1-0)^4}=27$$



                        as $xtoinfty$, so your result is correct.







                        share|cite|improve this answer












                        share|cite|improve this answer



                        share|cite|improve this answer










                        answered Dec 5 '18 at 5:15









                        MasacrosoMasacroso

                        13.1k41747




                        13.1k41747























                            3












                            $begingroup$

                            Yes, your answer is correct. An easy way is



                            begin{align}
                            lim_{x to ∞} f(x) &=lim_{x to ∞} frac{(3x+1)^3(x-1)}{(x-2)^4}\
                            &= lim_{x to ∞}frac{x^4cdot (3+frac{1}{x})^3(1-frac{1}{x})}{x^4cdot (1-frac{2}{x})^4}\
                            &= lim_{x to ∞} frac{(3+frac{1}{x})^3(1-frac{1}{x})}{ (1-frac{2}{x})^4}\
                            &=27
                            end{align}






                            share|cite|improve this answer









                            $endgroup$


















                              3












                              $begingroup$

                              Yes, your answer is correct. An easy way is



                              begin{align}
                              lim_{x to ∞} f(x) &=lim_{x to ∞} frac{(3x+1)^3(x-1)}{(x-2)^4}\
                              &= lim_{x to ∞}frac{x^4cdot (3+frac{1}{x})^3(1-frac{1}{x})}{x^4cdot (1-frac{2}{x})^4}\
                              &= lim_{x to ∞} frac{(3+frac{1}{x})^3(1-frac{1}{x})}{ (1-frac{2}{x})^4}\
                              &=27
                              end{align}






                              share|cite|improve this answer









                              $endgroup$
















                                3












                                3








                                3





                                $begingroup$

                                Yes, your answer is correct. An easy way is



                                begin{align}
                                lim_{x to ∞} f(x) &=lim_{x to ∞} frac{(3x+1)^3(x-1)}{(x-2)^4}\
                                &= lim_{x to ∞}frac{x^4cdot (3+frac{1}{x})^3(1-frac{1}{x})}{x^4cdot (1-frac{2}{x})^4}\
                                &= lim_{x to ∞} frac{(3+frac{1}{x})^3(1-frac{1}{x})}{ (1-frac{2}{x})^4}\
                                &=27
                                end{align}






                                share|cite|improve this answer









                                $endgroup$



                                Yes, your answer is correct. An easy way is



                                begin{align}
                                lim_{x to ∞} f(x) &=lim_{x to ∞} frac{(3x+1)^3(x-1)}{(x-2)^4}\
                                &= lim_{x to ∞}frac{x^4cdot (3+frac{1}{x})^3(1-frac{1}{x})}{x^4cdot (1-frac{2}{x})^4}\
                                &= lim_{x to ∞} frac{(3+frac{1}{x})^3(1-frac{1}{x})}{ (1-frac{2}{x})^4}\
                                &=27
                                end{align}







                                share|cite|improve this answer












                                share|cite|improve this answer



                                share|cite|improve this answer










                                answered Dec 5 '18 at 5:17









                                Thomas ShelbyThomas Shelby

                                3,7192525




                                3,7192525























                                    2












                                    $begingroup$

                                    You don't need to really check too much. It's sufficient to see that:



                                    $$(3x+1)^3(x-1) = 3^3x^4 + text{(other stuff of a lesser degree)}$$



                                    and



                                    $$(x-2)^4 = x^4 + text{(other stuff of a lesser degree)}$$



                                    Since the numerator and denominator have the same degree, the limit as $xto infty$ will be the ratio of the highest-degree terms' coefficients. Since the coefficients of the $x^4$ terms, respectively, are $3^3 = 27$ and $1$, then the limit evaluates to



                                    $$lim_{x to ∞} f(x) = frac{(3x+1)^3(x-1)}{(x-2)^4} = frac{27}{1} = 27$$



                                    So, tl;dr version, you're right. The argument could be made that you did a bit more work than necessary by multiplying everything out, but that's more an issue of personal taste in a case like this (since it's not a ton more work).






                                    share|cite|improve this answer









                                    $endgroup$


















                                      2












                                      $begingroup$

                                      You don't need to really check too much. It's sufficient to see that:



                                      $$(3x+1)^3(x-1) = 3^3x^4 + text{(other stuff of a lesser degree)}$$



                                      and



                                      $$(x-2)^4 = x^4 + text{(other stuff of a lesser degree)}$$



                                      Since the numerator and denominator have the same degree, the limit as $xto infty$ will be the ratio of the highest-degree terms' coefficients. Since the coefficients of the $x^4$ terms, respectively, are $3^3 = 27$ and $1$, then the limit evaluates to



                                      $$lim_{x to ∞} f(x) = frac{(3x+1)^3(x-1)}{(x-2)^4} = frac{27}{1} = 27$$



                                      So, tl;dr version, you're right. The argument could be made that you did a bit more work than necessary by multiplying everything out, but that's more an issue of personal taste in a case like this (since it's not a ton more work).






                                      share|cite|improve this answer









                                      $endgroup$
















                                        2












                                        2








                                        2





                                        $begingroup$

                                        You don't need to really check too much. It's sufficient to see that:



                                        $$(3x+1)^3(x-1) = 3^3x^4 + text{(other stuff of a lesser degree)}$$



                                        and



                                        $$(x-2)^4 = x^4 + text{(other stuff of a lesser degree)}$$



                                        Since the numerator and denominator have the same degree, the limit as $xto infty$ will be the ratio of the highest-degree terms' coefficients. Since the coefficients of the $x^4$ terms, respectively, are $3^3 = 27$ and $1$, then the limit evaluates to



                                        $$lim_{x to ∞} f(x) = frac{(3x+1)^3(x-1)}{(x-2)^4} = frac{27}{1} = 27$$



                                        So, tl;dr version, you're right. The argument could be made that you did a bit more work than necessary by multiplying everything out, but that's more an issue of personal taste in a case like this (since it's not a ton more work).






                                        share|cite|improve this answer









                                        $endgroup$



                                        You don't need to really check too much. It's sufficient to see that:



                                        $$(3x+1)^3(x-1) = 3^3x^4 + text{(other stuff of a lesser degree)}$$



                                        and



                                        $$(x-2)^4 = x^4 + text{(other stuff of a lesser degree)}$$



                                        Since the numerator and denominator have the same degree, the limit as $xto infty$ will be the ratio of the highest-degree terms' coefficients. Since the coefficients of the $x^4$ terms, respectively, are $3^3 = 27$ and $1$, then the limit evaluates to



                                        $$lim_{x to ∞} f(x) = frac{(3x+1)^3(x-1)}{(x-2)^4} = frac{27}{1} = 27$$



                                        So, tl;dr version, you're right. The argument could be made that you did a bit more work than necessary by multiplying everything out, but that's more an issue of personal taste in a case like this (since it's not a ton more work).







                                        share|cite|improve this answer












                                        share|cite|improve this answer



                                        share|cite|improve this answer










                                        answered Dec 5 '18 at 5:12









                                        Eevee TrainerEevee Trainer

                                        7,02311337




                                        7,02311337























                                            1












                                            $begingroup$

                                            Only to check or guess the possible result we can use that the constant term are negligeble then



                                            $$frac{(3x+1)^3(x-1)}{(x-2)^4}sim frac{(3x)^3(x)}{(x)^4}=frac{27x^4}{x^4}=27$$



                                            and then proceed for a rigorous proof as already indicated.






                                            share|cite|improve this answer









                                            $endgroup$


















                                              1












                                              $begingroup$

                                              Only to check or guess the possible result we can use that the constant term are negligeble then



                                              $$frac{(3x+1)^3(x-1)}{(x-2)^4}sim frac{(3x)^3(x)}{(x)^4}=frac{27x^4}{x^4}=27$$



                                              and then proceed for a rigorous proof as already indicated.






                                              share|cite|improve this answer









                                              $endgroup$
















                                                1












                                                1








                                                1





                                                $begingroup$

                                                Only to check or guess the possible result we can use that the constant term are negligeble then



                                                $$frac{(3x+1)^3(x-1)}{(x-2)^4}sim frac{(3x)^3(x)}{(x)^4}=frac{27x^4}{x^4}=27$$



                                                and then proceed for a rigorous proof as already indicated.






                                                share|cite|improve this answer









                                                $endgroup$



                                                Only to check or guess the possible result we can use that the constant term are negligeble then



                                                $$frac{(3x+1)^3(x-1)}{(x-2)^4}sim frac{(3x)^3(x)}{(x)^4}=frac{27x^4}{x^4}=27$$



                                                and then proceed for a rigorous proof as already indicated.







                                                share|cite|improve this answer












                                                share|cite|improve this answer



                                                share|cite|improve this answer










                                                answered Dec 5 '18 at 7:59









                                                gimusigimusi

                                                92.9k84494




                                                92.9k84494























                                                    1












                                                    $begingroup$

                                                    As an alternative method (for your practicing) you can rearrange degrees:
                                                    $$lim_{x to ∞} frac{(3x+1)^3(x-1)}{(x-2)^4}=lim_{x to ∞} left(frac{3x+1}{x-2}right)^3left(frac{x-1}{x-2}right)=cdots=3^3cdot 1=27.$$
                                                    Note: In the dots you should use the properies of limits.






                                                    share|cite|improve this answer









                                                    $endgroup$


















                                                      1












                                                      $begingroup$

                                                      As an alternative method (for your practicing) you can rearrange degrees:
                                                      $$lim_{x to ∞} frac{(3x+1)^3(x-1)}{(x-2)^4}=lim_{x to ∞} left(frac{3x+1}{x-2}right)^3left(frac{x-1}{x-2}right)=cdots=3^3cdot 1=27.$$
                                                      Note: In the dots you should use the properies of limits.






                                                      share|cite|improve this answer









                                                      $endgroup$
















                                                        1












                                                        1








                                                        1





                                                        $begingroup$

                                                        As an alternative method (for your practicing) you can rearrange degrees:
                                                        $$lim_{x to ∞} frac{(3x+1)^3(x-1)}{(x-2)^4}=lim_{x to ∞} left(frac{3x+1}{x-2}right)^3left(frac{x-1}{x-2}right)=cdots=3^3cdot 1=27.$$
                                                        Note: In the dots you should use the properies of limits.






                                                        share|cite|improve this answer









                                                        $endgroup$



                                                        As an alternative method (for your practicing) you can rearrange degrees:
                                                        $$lim_{x to ∞} frac{(3x+1)^3(x-1)}{(x-2)^4}=lim_{x to ∞} left(frac{3x+1}{x-2}right)^3left(frac{x-1}{x-2}right)=cdots=3^3cdot 1=27.$$
                                                        Note: In the dots you should use the properies of limits.







                                                        share|cite|improve this answer












                                                        share|cite|improve this answer



                                                        share|cite|improve this answer










                                                        answered Dec 5 '18 at 9:32









                                                        farruhotafarruhota

                                                        20.5k2740




                                                        20.5k2740






























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                                                            Biblatex bibliography style without URLs when DOI exists (in Overleaf with Zotero bibliography)

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                                                            2 2 I have a Zotero bibliography in Overleaf. I use biblatex . usepackage[backend=biber,citestyle=authoryear]{biblatex} addbibresource{zoteroALH.bib} Bibliography entries look like: @book{de_saint-gervais_uniformization_2016, title = {Uniformization of {Riemann} {Surfaces}}, isbn = {978-3-03719-145-3}, url = {https://www.ems-ph.org/books/book.php?proj_nr=198&srch=series%7Chem}, urldate = {2019-02-03}, author = {de Saint-Gervais, Paul Henri}, year = {2016}, doi = {10.4171/145} } In the bibliography down the paper the reference shows like this: There is the DOI, great, but also the URL which is unnecessary when the DOI is there. How can I remove the URL when the DOI is present? biblatex bibliographies overleaf zote
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                                                            ComboBox Display Member on multiple fields

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                                                            2 I'm trying to set the .DisplayMember property of a ComboBox in C#, but I want to bind it to multiple columns in the .DataSouce . My SQL looks like this: SELECT PersNbr, PersFirstName, PersMiddleName, PersLastName FROM Pers WHERE PersNbr = :persNbr; I'm saving this query in a DataTable so each column selected has it's own column in the Datatable . I want to make the .DisplayMember a combination of PersFirstName + PersMiddleName + PersLastName so their full name appears like this: comboBox.DisplayMemeber = "PersFirstName" + "PersMiddleName" + "PersLastName" I know I can just to this on the query: SELECT PersNbr, (PersFirstName || PersMiddleName || PersLastName) PersName and then just do this: comboBox.DisplayMember = "PersName"; but I
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                                                            Is it possible to collect Nectar points via Trainline?

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                                                            2 Since I am not eligible for any Railcard, the only available discount/bonus scheme is Nectar Card for me. GWR, SWR, LNER and Virgin Trains allow to collect Nectar points and I have already linked a card to my GWR account. A friend suggested me to use Trainline App, because it can search for bus tickets either. I had a look at the application, but could not see any option for Nectar. Is there a way to collect Nectar points when I buy tickets via Trainline? uk trains loyalty-programs trainline share | improve this question asked Dec 2 '18 at 15:50 ahmedus 3,224 5 19 49
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