Compute the limit $lim_{xto1}{frac{2x^3-2x^2+x-1}{x^3-x^2+3x-3}}$












0












$begingroup$


$$lim_{xto 1}{frac{2x^3-2x^2+x-1}{x^3-x^2+3x-3}}$$



What I tried: divided by $x^3$



$$lim_{xto1}{frac{2-frac{2}{x}+frac{1}{x^2}-frac{1}{x^3}}{1-frac{1}{x}+frac{3}{x^2}-frac{3}{x^3}}}$$



Then I plug in $x=1$



$$frac{2-2+1-1}{1-1+3-3}=frac{0}{0}$$



This is not correct, where did I make my mistake?






calculus limits rational-functions







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$endgroup$

















    0












    $begingroup$


    $$lim_{xto 1}{frac{2x^3-2x^2+x-1}{x^3-x^2+3x-3}}$$



    What I tried: divided by $x^3$



    $$lim_{xto1}{frac{2-frac{2}{x}+frac{1}{x^2}-frac{1}{x^3}}{1-frac{1}{x}+frac{3}{x^2}-frac{3}{x^3}}}$$



    Then I plug in $x=1$



    $$frac{2-2+1-1}{1-1+3-3}=frac{0}{0}$$



    This is not correct, where did I make my mistake?






    calculus limits rational-functions







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      $$lim_{xto 1}{frac{2x^3-2x^2+x-1}{x^3-x^2+3x-3}}$$



      What I tried: divided by $x^3$



      $$lim_{xto1}{frac{2-frac{2}{x}+frac{1}{x^2}-frac{1}{x^3}}{1-frac{1}{x}+frac{3}{x^2}-frac{3}{x^3}}}$$



      Then I plug in $x=1$



      $$frac{2-2+1-1}{1-1+3-3}=frac{0}{0}$$



      This is not correct, where did I make my mistake?






      calculus limits rational-functions







      share|cite|improve this question











      $endgroup$




      $$lim_{xto 1}{frac{2x^3-2x^2+x-1}{x^3-x^2+3x-3}}$$



      What I tried: divided by $x^3$



      $$lim_{xto1}{frac{2-frac{2}{x}+frac{1}{x^2}-frac{1}{x^3}}{1-frac{1}{x}+frac{3}{x^2}-frac{3}{x^3}}}$$



      Then I plug in $x=1$



      $$frac{2-2+1-1}{1-1+3-3}=frac{0}{0}$$



      This is not correct, where did I make my mistake?






      calculus limits rational-functions




      calculus limits rational-functions






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 8 '18 at 14:26









      Rodrigo de Azevedo

      12.9k41857




      12.9k41857










      asked Dec 8 '18 at 14:20









      Bili DebiliBili Debili

      1428




      1428






















          3 Answers
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          5












          $begingroup$

          We only evaluate it directly if we do not end up with indeterminate form.



          Hint:



          $$lim_{xto 1}{frac{2x^3-2x^2+x-1}{x^3-x^2+3x-3}}= lim_{xto 1}{frac{2x^2(x-1)+(x-1)}{x^2(x-1)+3(x-1)}}$$






          share|cite|improve this answer









          $endgroup$





















            4












            $begingroup$

            Since it looks like $0/0$, it means that both the denominator and numerator are zero. That means that $1$ is a root of both. What that tells you is that you should take $(x-1)$ as a factor, then simplify the fraction.






            share|cite|improve this answer









            $endgroup$





















              1












              $begingroup$

              That's not an effective way to solve since the limit is at $x=1$ and not to $infty$, indeed in that latter case your method would be fine since we obtain



              $$lim_{xtoinfty}{frac{2-frac{2}{x}+frac{1}{x^2}-frac{1}{x^3}}{1-frac{1}{x}+frac{3}{x^2}-frac{3}{x^3}}}=2$$



              As an effective alternative, in this case, we could use l'Hopital for example.




              Edit for more detail requested



              By l'Hopital we obtain



              $$lim_{xto 1}{frac{2x^3-2x^2+x-1}{x^3-x^2+3x-3}}stackrel{H.R.}=lim_{xto 1}{frac{6x^2-4x+1}{3x^2-2x+3}}=ldots$$



              Can you conclude?






              share|cite|improve this answer











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                3 Answers
                3






                active

                oldest

                votes








                3 Answers
                3






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                5












                $begingroup$

                We only evaluate it directly if we do not end up with indeterminate form.



                Hint:



                $$lim_{xto 1}{frac{2x^3-2x^2+x-1}{x^3-x^2+3x-3}}= lim_{xto 1}{frac{2x^2(x-1)+(x-1)}{x^2(x-1)+3(x-1)}}$$






                share|cite|improve this answer









                $endgroup$


















                  5












                  $begingroup$

                  We only evaluate it directly if we do not end up with indeterminate form.



                  Hint:



                  $$lim_{xto 1}{frac{2x^3-2x^2+x-1}{x^3-x^2+3x-3}}= lim_{xto 1}{frac{2x^2(x-1)+(x-1)}{x^2(x-1)+3(x-1)}}$$






                  share|cite|improve this answer









                  $endgroup$
















                    5












                    5








                    5





                    $begingroup$

                    We only evaluate it directly if we do not end up with indeterminate form.



                    Hint:



                    $$lim_{xto 1}{frac{2x^3-2x^2+x-1}{x^3-x^2+3x-3}}= lim_{xto 1}{frac{2x^2(x-1)+(x-1)}{x^2(x-1)+3(x-1)}}$$






                    share|cite|improve this answer









                    $endgroup$



                    We only evaluate it directly if we do not end up with indeterminate form.



                    Hint:



                    $$lim_{xto 1}{frac{2x^3-2x^2+x-1}{x^3-x^2+3x-3}}= lim_{xto 1}{frac{2x^2(x-1)+(x-1)}{x^2(x-1)+3(x-1)}}$$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 8 '18 at 14:24









                    Siong Thye GohSiong Thye Goh

                    101k1466117




                    101k1466117























                        4












                        $begingroup$

                        Since it looks like $0/0$, it means that both the denominator and numerator are zero. That means that $1$ is a root of both. What that tells you is that you should take $(x-1)$ as a factor, then simplify the fraction.






                        share|cite|improve this answer









                        $endgroup$


















                          4












                          $begingroup$

                          Since it looks like $0/0$, it means that both the denominator and numerator are zero. That means that $1$ is a root of both. What that tells you is that you should take $(x-1)$ as a factor, then simplify the fraction.






                          share|cite|improve this answer









                          $endgroup$
















                            4












                            4








                            4





                            $begingroup$

                            Since it looks like $0/0$, it means that both the denominator and numerator are zero. That means that $1$ is a root of both. What that tells you is that you should take $(x-1)$ as a factor, then simplify the fraction.






                            share|cite|improve this answer









                            $endgroup$



                            Since it looks like $0/0$, it means that both the denominator and numerator are zero. That means that $1$ is a root of both. What that tells you is that you should take $(x-1)$ as a factor, then simplify the fraction.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 8 '18 at 14:26









                            AndreiAndrei

                            11.8k21026




                            11.8k21026























                                1












                                $begingroup$

                                That's not an effective way to solve since the limit is at $x=1$ and not to $infty$, indeed in that latter case your method would be fine since we obtain



                                $$lim_{xtoinfty}{frac{2-frac{2}{x}+frac{1}{x^2}-frac{1}{x^3}}{1-frac{1}{x}+frac{3}{x^2}-frac{3}{x^3}}}=2$$



                                As an effective alternative, in this case, we could use l'Hopital for example.




                                Edit for more detail requested



                                By l'Hopital we obtain



                                $$lim_{xto 1}{frac{2x^3-2x^2+x-1}{x^3-x^2+3x-3}}stackrel{H.R.}=lim_{xto 1}{frac{6x^2-4x+1}{3x^2-2x+3}}=ldots$$



                                Can you conclude?






                                share|cite|improve this answer











                                $endgroup$


















                                  1












                                  $begingroup$

                                  That's not an effective way to solve since the limit is at $x=1$ and not to $infty$, indeed in that latter case your method would be fine since we obtain



                                  $$lim_{xtoinfty}{frac{2-frac{2}{x}+frac{1}{x^2}-frac{1}{x^3}}{1-frac{1}{x}+frac{3}{x^2}-frac{3}{x^3}}}=2$$



                                  As an effective alternative, in this case, we could use l'Hopital for example.




                                  Edit for more detail requested



                                  By l'Hopital we obtain



                                  $$lim_{xto 1}{frac{2x^3-2x^2+x-1}{x^3-x^2+3x-3}}stackrel{H.R.}=lim_{xto 1}{frac{6x^2-4x+1}{3x^2-2x+3}}=ldots$$



                                  Can you conclude?






                                  share|cite|improve this answer











                                  $endgroup$
















                                    1












                                    1








                                    1





                                    $begingroup$

                                    That's not an effective way to solve since the limit is at $x=1$ and not to $infty$, indeed in that latter case your method would be fine since we obtain



                                    $$lim_{xtoinfty}{frac{2-frac{2}{x}+frac{1}{x^2}-frac{1}{x^3}}{1-frac{1}{x}+frac{3}{x^2}-frac{3}{x^3}}}=2$$



                                    As an effective alternative, in this case, we could use l'Hopital for example.




                                    Edit for more detail requested



                                    By l'Hopital we obtain



                                    $$lim_{xto 1}{frac{2x^3-2x^2+x-1}{x^3-x^2+3x-3}}stackrel{H.R.}=lim_{xto 1}{frac{6x^2-4x+1}{3x^2-2x+3}}=ldots$$



                                    Can you conclude?






                                    share|cite|improve this answer











                                    $endgroup$



                                    That's not an effective way to solve since the limit is at $x=1$ and not to $infty$, indeed in that latter case your method would be fine since we obtain



                                    $$lim_{xtoinfty}{frac{2-frac{2}{x}+frac{1}{x^2}-frac{1}{x^3}}{1-frac{1}{x}+frac{3}{x^2}-frac{3}{x^3}}}=2$$



                                    As an effective alternative, in this case, we could use l'Hopital for example.




                                    Edit for more detail requested



                                    By l'Hopital we obtain



                                    $$lim_{xto 1}{frac{2x^3-2x^2+x-1}{x^3-x^2+3x-3}}stackrel{H.R.}=lim_{xto 1}{frac{6x^2-4x+1}{3x^2-2x+3}}=ldots$$



                                    Can you conclude?







                                    share|cite|improve this answer














                                    share|cite|improve this answer



                                    share|cite|improve this answer








                                    edited Dec 8 '18 at 14:46

























                                    answered Dec 8 '18 at 14:25









                                    gimusigimusi

                                    92.8k84494




                                    92.8k84494






























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