Limit using l'Hopital's rule with logaritmus












0












$begingroup$



What is $lim_{xrightarrow0}(1+2sin x)^frac{1}{tan x}$?




Computing $lim_{xrightarrow0}log(1+2sin x)^frac{1}{tan x}=lim_{xrightarrow0}frac{log(1+2sin x)}{tan x}$



Here, where I can use l'Hopital's rule I get:



$lim_{xrightarrow0}frac{log(1+2sin x)}{tan x}=2$



But now my question is:



$lim_{xrightarrow0}(1+2sin x)^frac{1}{tan x}$ = $lim_{xrightarrow0}frac{log(1+2sin x)}{tan x}=2$?



I don't think they are the same, but can I remove the log? Maybe it's easier than I think, but now I don't know how to go on.






real-analysis limits







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$endgroup$

















    0












    $begingroup$



    What is $lim_{xrightarrow0}(1+2sin x)^frac{1}{tan x}$?




    Computing $lim_{xrightarrow0}log(1+2sin x)^frac{1}{tan x}=lim_{xrightarrow0}frac{log(1+2sin x)}{tan x}$



    Here, where I can use l'Hopital's rule I get:



    $lim_{xrightarrow0}frac{log(1+2sin x)}{tan x}=2$



    But now my question is:



    $lim_{xrightarrow0}(1+2sin x)^frac{1}{tan x}$ = $lim_{xrightarrow0}frac{log(1+2sin x)}{tan x}=2$?



    I don't think they are the same, but can I remove the log? Maybe it's easier than I think, but now I don't know how to go on.






    real-analysis limits







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$



      What is $lim_{xrightarrow0}(1+2sin x)^frac{1}{tan x}$?




      Computing $lim_{xrightarrow0}log(1+2sin x)^frac{1}{tan x}=lim_{xrightarrow0}frac{log(1+2sin x)}{tan x}$



      Here, where I can use l'Hopital's rule I get:



      $lim_{xrightarrow0}frac{log(1+2sin x)}{tan x}=2$



      But now my question is:



      $lim_{xrightarrow0}(1+2sin x)^frac{1}{tan x}$ = $lim_{xrightarrow0}frac{log(1+2sin x)}{tan x}=2$?



      I don't think they are the same, but can I remove the log? Maybe it's easier than I think, but now I don't know how to go on.






      real-analysis limits







      share|cite|improve this question











      $endgroup$





      What is $lim_{xrightarrow0}(1+2sin x)^frac{1}{tan x}$?




      Computing $lim_{xrightarrow0}log(1+2sin x)^frac{1}{tan x}=lim_{xrightarrow0}frac{log(1+2sin x)}{tan x}$



      Here, where I can use l'Hopital's rule I get:



      $lim_{xrightarrow0}frac{log(1+2sin x)}{tan x}=2$



      But now my question is:



      $lim_{xrightarrow0}(1+2sin x)^frac{1}{tan x}$ = $lim_{xrightarrow0}frac{log(1+2sin x)}{tan x}=2$?



      I don't think they are the same, but can I remove the log? Maybe it's easier than I think, but now I don't know how to go on.






      real-analysis limits




      real-analysis limits






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 29 '18 at 22:31









      gimusi

      92.8k84494




      92.8k84494










      asked Nov 29 '18 at 22:19









      DadaDada

      7510




      7510






















          3 Answers
          3






          active

          oldest

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          2












          $begingroup$

          You mean $$lim_{xto 0}(1+2sin x)^{tfrac{1}{tan x}}=explim_{xto 0}frac{ln (1+2sin x)}{tan x}=e^2.$$






          share|cite|improve this answer









          $endgroup$





















            2












            $begingroup$

            The first equality you write is incorrect. You can evaluate
            $$
            lim_{xto0}frac{log(1+2sin x)}{tan x}
            $$
            in order to compute the given limit, but they are not equal; if $l$ is the latter limit, then the one you're looking for is $e^l$.



            Now, with l'Hôpital,
            $$
            lim_{xto0}frac{log(1+2sin x)}{tan x}=
            lim_{xto0}frac{dfrac{2cos x}{1+2sin x}}{1+tan^2x}=2
            $$
            and therefore
            $$
            lim_{xto0}(1+2sin x)^{1/tan x}=e^2
            $$






            share|cite|improve this answer









            $endgroup$





















              2












              $begingroup$

              As an alternative



              $$large (1+2 sin x)^frac{1}{tan x}=left[(1+2 sin x)^{frac1{2sin x}}right]^frac{2sin x}{tan x} to e^2$$






              share|cite|improve this answer









              $endgroup$













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                3 Answers
                3






                active

                oldest

                votes








                3 Answers
                3






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                2












                $begingroup$

                You mean $$lim_{xto 0}(1+2sin x)^{tfrac{1}{tan x}}=explim_{xto 0}frac{ln (1+2sin x)}{tan x}=e^2.$$






                share|cite|improve this answer









                $endgroup$


















                  2












                  $begingroup$

                  You mean $$lim_{xto 0}(1+2sin x)^{tfrac{1}{tan x}}=explim_{xto 0}frac{ln (1+2sin x)}{tan x}=e^2.$$






                  share|cite|improve this answer









                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    You mean $$lim_{xto 0}(1+2sin x)^{tfrac{1}{tan x}}=explim_{xto 0}frac{ln (1+2sin x)}{tan x}=e^2.$$






                    share|cite|improve this answer









                    $endgroup$



                    You mean $$lim_{xto 0}(1+2sin x)^{tfrac{1}{tan x}}=explim_{xto 0}frac{ln (1+2sin x)}{tan x}=e^2.$$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Nov 29 '18 at 22:26









                    J.G.J.G.

                    25.7k22540




                    25.7k22540























                        2












                        $begingroup$

                        The first equality you write is incorrect. You can evaluate
                        $$
                        lim_{xto0}frac{log(1+2sin x)}{tan x}
                        $$
                        in order to compute the given limit, but they are not equal; if $l$ is the latter limit, then the one you're looking for is $e^l$.



                        Now, with l'Hôpital,
                        $$
                        lim_{xto0}frac{log(1+2sin x)}{tan x}=
                        lim_{xto0}frac{dfrac{2cos x}{1+2sin x}}{1+tan^2x}=2
                        $$
                        and therefore
                        $$
                        lim_{xto0}(1+2sin x)^{1/tan x}=e^2
                        $$






                        share|cite|improve this answer









                        $endgroup$


















                          2












                          $begingroup$

                          The first equality you write is incorrect. You can evaluate
                          $$
                          lim_{xto0}frac{log(1+2sin x)}{tan x}
                          $$
                          in order to compute the given limit, but they are not equal; if $l$ is the latter limit, then the one you're looking for is $e^l$.



                          Now, with l'Hôpital,
                          $$
                          lim_{xto0}frac{log(1+2sin x)}{tan x}=
                          lim_{xto0}frac{dfrac{2cos x}{1+2sin x}}{1+tan^2x}=2
                          $$
                          and therefore
                          $$
                          lim_{xto0}(1+2sin x)^{1/tan x}=e^2
                          $$






                          share|cite|improve this answer









                          $endgroup$
















                            2












                            2








                            2





                            $begingroup$

                            The first equality you write is incorrect. You can evaluate
                            $$
                            lim_{xto0}frac{log(1+2sin x)}{tan x}
                            $$
                            in order to compute the given limit, but they are not equal; if $l$ is the latter limit, then the one you're looking for is $e^l$.



                            Now, with l'Hôpital,
                            $$
                            lim_{xto0}frac{log(1+2sin x)}{tan x}=
                            lim_{xto0}frac{dfrac{2cos x}{1+2sin x}}{1+tan^2x}=2
                            $$
                            and therefore
                            $$
                            lim_{xto0}(1+2sin x)^{1/tan x}=e^2
                            $$






                            share|cite|improve this answer









                            $endgroup$



                            The first equality you write is incorrect. You can evaluate
                            $$
                            lim_{xto0}frac{log(1+2sin x)}{tan x}
                            $$
                            in order to compute the given limit, but they are not equal; if $l$ is the latter limit, then the one you're looking for is $e^l$.



                            Now, with l'Hôpital,
                            $$
                            lim_{xto0}frac{log(1+2sin x)}{tan x}=
                            lim_{xto0}frac{dfrac{2cos x}{1+2sin x}}{1+tan^2x}=2
                            $$
                            and therefore
                            $$
                            lim_{xto0}(1+2sin x)^{1/tan x}=e^2
                            $$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Nov 29 '18 at 22:26









                            egregegreg

                            181k1485203




                            181k1485203























                                2












                                $begingroup$

                                As an alternative



                                $$large (1+2 sin x)^frac{1}{tan x}=left[(1+2 sin x)^{frac1{2sin x}}right]^frac{2sin x}{tan x} to e^2$$






                                share|cite|improve this answer









                                $endgroup$


















                                  2












                                  $begingroup$

                                  As an alternative



                                  $$large (1+2 sin x)^frac{1}{tan x}=left[(1+2 sin x)^{frac1{2sin x}}right]^frac{2sin x}{tan x} to e^2$$






                                  share|cite|improve this answer









                                  $endgroup$
















                                    2












                                    2








                                    2





                                    $begingroup$

                                    As an alternative



                                    $$large (1+2 sin x)^frac{1}{tan x}=left[(1+2 sin x)^{frac1{2sin x}}right]^frac{2sin x}{tan x} to e^2$$






                                    share|cite|improve this answer









                                    $endgroup$



                                    As an alternative



                                    $$large (1+2 sin x)^frac{1}{tan x}=left[(1+2 sin x)^{frac1{2sin x}}right]^frac{2sin x}{tan x} to e^2$$







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Nov 29 '18 at 22:28









                                    gimusigimusi

                                    92.8k84494




                                    92.8k84494






























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