Finding limit as $x rightarrow 0$












3












$begingroup$


The question asks to find




$$lim _{x rightarrow 0} dfrac{sin x^n}{left(sin xright)^m} ;;;forall; m<n$$




I solved it by applying the L'Hopital rule since it is of the form $frac{0}{0}$ and then once I differentiated we get $$lim_{x rightarrow0} left(frac{frac{d sin x^n}{dx}}{frac{ left(d sin x right)^m}{dx}}right),
$$
which gives us
$$
lim_{x rightarrow0} frac{ncos x cdot x^{n-1}}{left(m sin xright)^{m-1} cos x}.
$$
Now this gives us a function of the form $$frac{n x^{n-1}}{left(m sin xright)^{m-1}},
$$
which means that since $n>m$ now, we can keep differentiating the denominator and we will eventually get something of the form $frac{0}{k}$ which means the limit must be $0$.
This is how I have gone about it. I have $2$ questions:
$1.$ Is there a better method than this to solve it?
$2.$ Can we find out the limit of this function if $m>n$?






calculus limits







share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    The question asks to find




    $$lim _{x rightarrow 0} dfrac{sin x^n}{left(sin xright)^m} ;;;forall; m<n$$




    I solved it by applying the L'Hopital rule since it is of the form $frac{0}{0}$ and then once I differentiated we get $$lim_{x rightarrow0} left(frac{frac{d sin x^n}{dx}}{frac{ left(d sin x right)^m}{dx}}right),
    $$
    which gives us
    $$
    lim_{x rightarrow0} frac{ncos x cdot x^{n-1}}{left(m sin xright)^{m-1} cos x}.
    $$
    Now this gives us a function of the form $$frac{n x^{n-1}}{left(m sin xright)^{m-1}},
    $$
    which means that since $n>m$ now, we can keep differentiating the denominator and we will eventually get something of the form $frac{0}{k}$ which means the limit must be $0$.
    This is how I have gone about it. I have $2$ questions:
    $1.$ Is there a better method than this to solve it?
    $2.$ Can we find out the limit of this function if $m>n$?






    calculus limits







    share|cite|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$


      The question asks to find




      $$lim _{x rightarrow 0} dfrac{sin x^n}{left(sin xright)^m} ;;;forall; m<n$$




      I solved it by applying the L'Hopital rule since it is of the form $frac{0}{0}$ and then once I differentiated we get $$lim_{x rightarrow0} left(frac{frac{d sin x^n}{dx}}{frac{ left(d sin x right)^m}{dx}}right),
      $$
      which gives us
      $$
      lim_{x rightarrow0} frac{ncos x cdot x^{n-1}}{left(m sin xright)^{m-1} cos x}.
      $$
      Now this gives us a function of the form $$frac{n x^{n-1}}{left(m sin xright)^{m-1}},
      $$
      which means that since $n>m$ now, we can keep differentiating the denominator and we will eventually get something of the form $frac{0}{k}$ which means the limit must be $0$.
      This is how I have gone about it. I have $2$ questions:
      $1.$ Is there a better method than this to solve it?
      $2.$ Can we find out the limit of this function if $m>n$?






      calculus limits







      share|cite|improve this question











      $endgroup$




      The question asks to find




      $$lim _{x rightarrow 0} dfrac{sin x^n}{left(sin xright)^m} ;;;forall; m<n$$




      I solved it by applying the L'Hopital rule since it is of the form $frac{0}{0}$ and then once I differentiated we get $$lim_{x rightarrow0} left(frac{frac{d sin x^n}{dx}}{frac{ left(d sin x right)^m}{dx}}right),
      $$
      which gives us
      $$
      lim_{x rightarrow0} frac{ncos x cdot x^{n-1}}{left(m sin xright)^{m-1} cos x}.
      $$
      Now this gives us a function of the form $$frac{n x^{n-1}}{left(m sin xright)^{m-1}},
      $$
      which means that since $n>m$ now, we can keep differentiating the denominator and we will eventually get something of the form $frac{0}{k}$ which means the limit must be $0$.
      This is how I have gone about it. I have $2$ questions:
      $1.$ Is there a better method than this to solve it?
      $2.$ Can we find out the limit of this function if $m>n$?






      calculus limits




      calculus limits






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 26 at 8:35









      Mee Seong Im

      2,8151617




      2,8151617










      asked Jan 26 at 8:20









      Prakhar NagpalPrakhar Nagpal

      752318




      752318






















          2 Answers
          2






          active

          oldest

          votes


















          7












          $begingroup$

          Write the function as follows:



          $$ frac{ sin x^n }{ x^n } cdot frac{x^m}{sin^m x} cdot x^{n-m}$$



          Now, in the limit we obtain $1 cdot 1^m cdot 0 = 0 $ since $n-m>0$. We have used the elementary fact that



          $$ lim_{x to 0} frac{ sin x }{x} = 1 $$






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$


            $$lim _{x rightarrow 0} dfrac{sin x^n}{left(sin xright)^m} ;;;forall; m<n$$




            $$= lim limits_{x rightarrow 0} frac{x^n (1 - frac{x^{2n}}{3!}dots)}{x^m[ 1 - frac{x^2}{3!} dots]^m} = limlimits_{x rightarrow0} x^{n-m} =0$$



            By the way, $sin x = x = frac {x^3}{3!} + frac{x^5}{5!} dots$



            and $cos x = 1 - frac {x^2}{2!} + frac{x^4}{4!} dots$






            share|cite|improve this answer











            $endgroup$













              Your Answer





              StackExchange.ifUsing("editor", function () {
              return StackExchange.using("mathjaxEditing", function () {
              StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
              StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
              });
              });
              }, "mathjax-editing");

              StackExchange.ready(function() {
              var channelOptions = {
              tags: "".split(" "),
              id: "69"
              };
              initTagRenderer("".split(" "), "".split(" "), channelOptions);

              StackExchange.using("externalEditor", function() {
              // Have to fire editor after snippets, if snippets enabled
              if (StackExchange.settings.snippets.snippetsEnabled) {
              StackExchange.using("snippets", function() {
              createEditor();
              });
              }
              else {
              createEditor();
              }
              });

              function createEditor() {
              StackExchange.prepareEditor({
              heartbeatType: 'answer',
              autoActivateHeartbeat: false,
              convertImagesToLinks: true,
              noModals: true,
              showLowRepImageUploadWarning: true,
              reputationToPostImages: 10,
              bindNavPrevention: true,
              postfix: "",
              imageUploader: {
              brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
              contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
              allowUrls: true
              },
              noCode: true, onDemand: true,
              discardSelector: ".discard-answer"
              ,immediatelyShowMarkdownHelp:true
              });


              }
              });














              draft saved

              draft discarded


















              StackExchange.ready(
              function () {
              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3088018%2ffinding-limit-as-x-rightarrow-0%23new-answer', 'question_page');
              }
              );

              Post as a guest















              Required, but never shown

























              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              7












              $begingroup$

              Write the function as follows:



              $$ frac{ sin x^n }{ x^n } cdot frac{x^m}{sin^m x} cdot x^{n-m}$$



              Now, in the limit we obtain $1 cdot 1^m cdot 0 = 0 $ since $n-m>0$. We have used the elementary fact that



              $$ lim_{x to 0} frac{ sin x }{x} = 1 $$






              share|cite|improve this answer









              $endgroup$


















                7












                $begingroup$

                Write the function as follows:



                $$ frac{ sin x^n }{ x^n } cdot frac{x^m}{sin^m x} cdot x^{n-m}$$



                Now, in the limit we obtain $1 cdot 1^m cdot 0 = 0 $ since $n-m>0$. We have used the elementary fact that



                $$ lim_{x to 0} frac{ sin x }{x} = 1 $$






                share|cite|improve this answer









                $endgroup$
















                  7












                  7








                  7





                  $begingroup$

                  Write the function as follows:



                  $$ frac{ sin x^n }{ x^n } cdot frac{x^m}{sin^m x} cdot x^{n-m}$$



                  Now, in the limit we obtain $1 cdot 1^m cdot 0 = 0 $ since $n-m>0$. We have used the elementary fact that



                  $$ lim_{x to 0} frac{ sin x }{x} = 1 $$






                  share|cite|improve this answer









                  $endgroup$



                  Write the function as follows:



                  $$ frac{ sin x^n }{ x^n } cdot frac{x^m}{sin^m x} cdot x^{n-m}$$



                  Now, in the limit we obtain $1 cdot 1^m cdot 0 = 0 $ since $n-m>0$. We have used the elementary fact that



                  $$ lim_{x to 0} frac{ sin x }{x} = 1 $$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 26 at 8:24









                  Jimmy SabaterJimmy Sabater

                  3,054325




                  3,054325























                      1












                      $begingroup$


                      $$lim _{x rightarrow 0} dfrac{sin x^n}{left(sin xright)^m} ;;;forall; m<n$$




                      $$= lim limits_{x rightarrow 0} frac{x^n (1 - frac{x^{2n}}{3!}dots)}{x^m[ 1 - frac{x^2}{3!} dots]^m} = limlimits_{x rightarrow0} x^{n-m} =0$$



                      By the way, $sin x = x = frac {x^3}{3!} + frac{x^5}{5!} dots$



                      and $cos x = 1 - frac {x^2}{2!} + frac{x^4}{4!} dots$






                      share|cite|improve this answer











                      $endgroup$


















                        1












                        $begingroup$


                        $$lim _{x rightarrow 0} dfrac{sin x^n}{left(sin xright)^m} ;;;forall; m<n$$




                        $$= lim limits_{x rightarrow 0} frac{x^n (1 - frac{x^{2n}}{3!}dots)}{x^m[ 1 - frac{x^2}{3!} dots]^m} = limlimits_{x rightarrow0} x^{n-m} =0$$



                        By the way, $sin x = x = frac {x^3}{3!} + frac{x^5}{5!} dots$



                        and $cos x = 1 - frac {x^2}{2!} + frac{x^4}{4!} dots$






                        share|cite|improve this answer











                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$


                          $$lim _{x rightarrow 0} dfrac{sin x^n}{left(sin xright)^m} ;;;forall; m<n$$




                          $$= lim limits_{x rightarrow 0} frac{x^n (1 - frac{x^{2n}}{3!}dots)}{x^m[ 1 - frac{x^2}{3!} dots]^m} = limlimits_{x rightarrow0} x^{n-m} =0$$



                          By the way, $sin x = x = frac {x^3}{3!} + frac{x^5}{5!} dots$



                          and $cos x = 1 - frac {x^2}{2!} + frac{x^4}{4!} dots$






                          share|cite|improve this answer











                          $endgroup$




                          $$lim _{x rightarrow 0} dfrac{sin x^n}{left(sin xright)^m} ;;;forall; m<n$$




                          $$= lim limits_{x rightarrow 0} frac{x^n (1 - frac{x^{2n}}{3!}dots)}{x^m[ 1 - frac{x^2}{3!} dots]^m} = limlimits_{x rightarrow0} x^{n-m} =0$$



                          By the way, $sin x = x = frac {x^3}{3!} + frac{x^5}{5!} dots$



                          and $cos x = 1 - frac {x^2}{2!} + frac{x^4}{4!} dots$







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Jan 27 at 12:51

























                          answered Jan 27 at 12:45









                          Abhas Kumar SinhaAbhas Kumar Sinha

                          304115




                          304115






























                              draft saved

                              draft discarded




















































                              Thanks for contributing an answer to Mathematics Stack Exchange!


                              • Please be sure to answer the question. Provide details and share your research!

                              But avoid …



                              • Asking for help, clarification, or responding to other answers.

                              • Making statements based on opinion; back them up with references or personal experience.


                              Use MathJax to format equations. MathJax reference.


                              To learn more, see our tips on writing great answers.




                              draft saved


                              draft discarded














                              StackExchange.ready(
                              function () {
                              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3088018%2ffinding-limit-as-x-rightarrow-0%23new-answer', 'question_page');
                              }
                              );

                              Post as a guest















                              Required, but never shown





















































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown

































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown







                              Popular posts from this blog

                              Can a sorcerer learn a 5th-level spell early by creating spell slots using the Font of Magic feature?

                              Image
                              28 6 $begingroup$ Per the Font of Magic feature, sorcerer can use Flexible Casting to create 5th-level spell slots at level 7, even though they are not typically available until level 9. You can transform unexpended sorcery points into one spell slot as a bonus action on your turn. The Creating Spell Slots table shows the cost of creating a spell slot of [5th level is 7] If such a sorcerer levels up while still having this 5th-level spell slot, can they choose a 5th-level spell as the spell they gain upon levelling up? The Spellcasting feature states: Additionally, when you gain a level in this class, you can choose one of the sorcerer spells you know and replace it with another spell from the sorcerer spell list, which also must be of a level for which you have spell slots.
                              Read more

                              Does disintegrating a polymorphed enemy still kill it after the 2018 errata?

                              Image
                              up vote 13 down vote favorite 1 After the 2018 errata, the disintegrate spell description now reads: A creature targeted by this spell must make a Dexterity saving throw. On a failed save, the target takes 10d6 + 40 force damage. The target is disintegrated if this damage leaves it with 0 hit points. If you were to polymorph an enemy into a rat and then disintegrate it, would the enemy be disintegrated or would it just return to its original form? I know that for Druids, it’s not an instant kill anymore, but is this the case for polymorph as well? dnd-5e spells polymorph errata share | improve this question edited 18 hours ago
                              Read more

                              A Topological Invariant for $pi_3(U(n))$

                              Image
                              3 3 $begingroup$ Recently, I saw a construction of topological invariant for $pi_3(U(n))$ with $ngeq 2$ : $$ N=frac{1}{24pi^2}int_{S^3} d^3x epsilon^{ijk} Tr[(U^{-1}partial_{x_i}U)(U^{-1}partial_{x_j}U)(U^{-1}partial_{x_k}U)] , $$ where $Uin U(n)$ depends on $boldsymbol{x}=(x_1,x_2,x_3)in S^3$ , $epsilon^{ijk}$ is the Levi-Civita symbol, $i,j,k=1,2,3$ , and the duplicated indexes are summed over. It is claimed that $N$ is an integer, but why? Update 02/02/2019 I think I got an argument for $n=2$ . In this case, $U=e^{i varphi} q$ with $qin SU(2)$ . Due to the trace and the Levi-Civita symbol in $N$ , $varphi$ does not contribute to $N$ . As $Tr[q^{dagger}partial_i q]=0$ and $(q^{dagger}partial_i q)^{dagger}=-q^{dagger}partial_i q$ , $q^{dagger}partial_i q$ in geneeral has the form $q^{dagger}partia
                              Read more