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
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          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









                  $endgroup$













                    Your Answer





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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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                                                            How to change which sound is reproduced for terminal bell?

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

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                                                            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:"
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                                                            Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

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                                                            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
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