Limlit of $frac{x-xln(1+x)-ln(1+x)}{x^3+x^2}$ at $0$ without l'Hôpital or Taylor series












4












$begingroup$


I have $f : x mapsto frac{ln(1+x)}{x}$ which derivative is $$frac{1}{x(1+x)}-frac{ln(1+x)}{x^2}$$
I want to find the limit as $x$ goes to $0$ of this derivative. I've tried simplifying the expression to $$frac{x - xln(1+x) - ln(1+x)}{x^3 + x^2}$$ but I'm still struggling. Also I can't use l'Hôpital or Taylor series. Any help would be appreciated, thanks in advance! (And sorry if I did anything wrong here)






logarithms limits-without-lhopital indeterminate-forms







share|cite|improve this question











$endgroup$

















    4












    $begingroup$


    I have $f : x mapsto frac{ln(1+x)}{x}$ which derivative is $$frac{1}{x(1+x)}-frac{ln(1+x)}{x^2}$$
    I want to find the limit as $x$ goes to $0$ of this derivative. I've tried simplifying the expression to $$frac{x - xln(1+x) - ln(1+x)}{x^3 + x^2}$$ but I'm still struggling. Also I can't use l'Hôpital or Taylor series. Any help would be appreciated, thanks in advance! (And sorry if I did anything wrong here)






    logarithms limits-without-lhopital indeterminate-forms







    share|cite|improve this question











    $endgroup$















      4












      4








      4


      1



      $begingroup$


      I have $f : x mapsto frac{ln(1+x)}{x}$ which derivative is $$frac{1}{x(1+x)}-frac{ln(1+x)}{x^2}$$
      I want to find the limit as $x$ goes to $0$ of this derivative. I've tried simplifying the expression to $$frac{x - xln(1+x) - ln(1+x)}{x^3 + x^2}$$ but I'm still struggling. Also I can't use l'Hôpital or Taylor series. Any help would be appreciated, thanks in advance! (And sorry if I did anything wrong here)






      logarithms limits-without-lhopital indeterminate-forms







      share|cite|improve this question











      $endgroup$




      I have $f : x mapsto frac{ln(1+x)}{x}$ which derivative is $$frac{1}{x(1+x)}-frac{ln(1+x)}{x^2}$$
      I want to find the limit as $x$ goes to $0$ of this derivative. I've tried simplifying the expression to $$frac{x - xln(1+x) - ln(1+x)}{x^3 + x^2}$$ but I'm still struggling. Also I can't use l'Hôpital or Taylor series. Any help would be appreciated, thanks in advance! (And sorry if I did anything wrong here)






      logarithms limits-without-lhopital indeterminate-forms




      logarithms limits-without-lhopital indeterminate-forms






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 5 at 0:58









      mrtaurho

      6,11271641




      6,11271641










      asked Jan 4 at 23:41









      HeryonHeryon

      214




      214






















          4 Answers
          4






          active

          oldest

          votes


















          3












          $begingroup$

          First, you can decompose the first term in your derivative as:



          $frac{x}{1+x} = frac{1}{x}-frac{1}{1+x}$.



          Second, you can expand the second term using Taylor series:



          $frac{ln(1+x)}{x^2}$



          $ = frac{1}{x^2} left( x - frac{x^2}{2} + frac{x^3}{3} - .. right)$



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



          Putting it together, your derivative is:



          $frac{1}{x}-frac{1}{1+x} - frac{1}{x} + frac{1}{2} - frac{x}{3} +...$




          • The $1/x$ terms cancel out.


          • $lim_{x rightarrow 0} frac{1}{1+x} = 1$.


          • $lim_{x rightarrow 0}$ of all terms with $x$ or higher powers of $x$ in the numerator is 0.


          So you are left with $-1 + frac{1}{2} = -frac{1}{2}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Clever, thanks. Any way to do this without Taylor series? (Just to know)
            $endgroup$
            – Heryon
            Jan 5 at 0:06



















          2












          $begingroup$

          Your original function $f(x)=frac{ln(1+x)}{x}$ can be written using the taylor series expansion of $ln(1+x)$ as follows:
          $$ln(1+x)=sum_{k=1}^{infty}frac{(-1)^{k+1}x^{k}}{k}=x-frac{x^2}{2}+frac{x^3}{3}-...$$
          $$thereforefrac{ln(1+x)}{x}=sum_{k=1}^{infty}frac{(-1)^{k+1}x^{k-1}}{k}=sum_{k=0}^{infty}frac{(-1)^{k}x^{k}}{k+1}=1-frac{x}{2}+frac{x^2}{3}-...$$
          $$frac{d}{dx}Bigg(frac{ln(1+x)}{x}Bigg)=frac{d}{dx}bigg(1-frac{x}{2}+frac{x^2}{3}-...bigg)=-frac{1}{2}+frac{2x}{3}-frac{3x^2}{4}+...$$
          So,
          $$lim_{xto0}Bigg(frac{d}{dx}Bigg(frac{ln(1+x)}{x}Bigg)Bigg)=lim_{xto0}bigg(-frac{1}{2}+frac{2x}{3}-frac{3x^2}{4}+...bigg)=fbox{$-frac{1}{2}$}$$






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            $frac {1}{x(x+1)} - frac {ln(x+1)}{x^2} = frac {d}{dx} frac {ln(x+1)}{x}$



            At which point I am inclined to use a series



            $frac {d}{dx} (1 - frac x2 + cdots)$



            as $x$ goes to $0$ equals $-frac 12$



            Without using a series, I have.



            $frac {d}{dx} frac {ln(x+1)}{x}$ evaluated at $x = 0$



            is $lim_limits{hto 0} frac{ ln(1+h) - 1}{h^2}$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Yes I can't use that too, edited the post, but thanks!
              $endgroup$
              – Heryon
              Jan 5 at 0:33



















            0












            $begingroup$

            Because $frac{x}{1+x}<ln(1+x)<x$



            So when $ x rightarrow 0+, f(x) rightarrow 1$



            When $xrightarrow0-$ is similar






            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%2f3062245%2flimlit-of-fracx-x-ln1x-ln1xx3x2-at-0-without-lh%25c3%25b4pital-or-tay%23new-answer', 'question_page');
              }
              );

              Post as a guest















              Required, but never shown

























              4 Answers
              4






              active

              oldest

              votes








              4 Answers
              4






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              3












              $begingroup$

              First, you can decompose the first term in your derivative as:



              $frac{x}{1+x} = frac{1}{x}-frac{1}{1+x}$.



              Second, you can expand the second term using Taylor series:



              $frac{ln(1+x)}{x^2}$



              $ = frac{1}{x^2} left( x - frac{x^2}{2} + frac{x^3}{3} - .. right)$



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



              Putting it together, your derivative is:



              $frac{1}{x}-frac{1}{1+x} - frac{1}{x} + frac{1}{2} - frac{x}{3} +...$




              • The $1/x$ terms cancel out.


              • $lim_{x rightarrow 0} frac{1}{1+x} = 1$.


              • $lim_{x rightarrow 0}$ of all terms with $x$ or higher powers of $x$ in the numerator is 0.


              So you are left with $-1 + frac{1}{2} = -frac{1}{2}$.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Clever, thanks. Any way to do this without Taylor series? (Just to know)
                $endgroup$
                – Heryon
                Jan 5 at 0:06
















              3












              $begingroup$

              First, you can decompose the first term in your derivative as:



              $frac{x}{1+x} = frac{1}{x}-frac{1}{1+x}$.



              Second, you can expand the second term using Taylor series:



              $frac{ln(1+x)}{x^2}$



              $ = frac{1}{x^2} left( x - frac{x^2}{2} + frac{x^3}{3} - .. right)$



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



              Putting it together, your derivative is:



              $frac{1}{x}-frac{1}{1+x} - frac{1}{x} + frac{1}{2} - frac{x}{3} +...$




              • The $1/x$ terms cancel out.


              • $lim_{x rightarrow 0} frac{1}{1+x} = 1$.


              • $lim_{x rightarrow 0}$ of all terms with $x$ or higher powers of $x$ in the numerator is 0.


              So you are left with $-1 + frac{1}{2} = -frac{1}{2}$.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Clever, thanks. Any way to do this without Taylor series? (Just to know)
                $endgroup$
                – Heryon
                Jan 5 at 0:06














              3












              3








              3





              $begingroup$

              First, you can decompose the first term in your derivative as:



              $frac{x}{1+x} = frac{1}{x}-frac{1}{1+x}$.



              Second, you can expand the second term using Taylor series:



              $frac{ln(1+x)}{x^2}$



              $ = frac{1}{x^2} left( x - frac{x^2}{2} + frac{x^3}{3} - .. right)$



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



              Putting it together, your derivative is:



              $frac{1}{x}-frac{1}{1+x} - frac{1}{x} + frac{1}{2} - frac{x}{3} +...$




              • The $1/x$ terms cancel out.


              • $lim_{x rightarrow 0} frac{1}{1+x} = 1$.


              • $lim_{x rightarrow 0}$ of all terms with $x$ or higher powers of $x$ in the numerator is 0.


              So you are left with $-1 + frac{1}{2} = -frac{1}{2}$.






              share|cite|improve this answer









              $endgroup$



              First, you can decompose the first term in your derivative as:



              $frac{x}{1+x} = frac{1}{x}-frac{1}{1+x}$.



              Second, you can expand the second term using Taylor series:



              $frac{ln(1+x)}{x^2}$



              $ = frac{1}{x^2} left( x - frac{x^2}{2} + frac{x^3}{3} - .. right)$



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



              Putting it together, your derivative is:



              $frac{1}{x}-frac{1}{1+x} - frac{1}{x} + frac{1}{2} - frac{x}{3} +...$




              • The $1/x$ terms cancel out.


              • $lim_{x rightarrow 0} frac{1}{1+x} = 1$.


              • $lim_{x rightarrow 0}$ of all terms with $x$ or higher powers of $x$ in the numerator is 0.


              So you are left with $-1 + frac{1}{2} = -frac{1}{2}$.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Jan 4 at 23:51









              Aditya DuaAditya Dua

              1,15418




              1,15418












              • $begingroup$
                Clever, thanks. Any way to do this without Taylor series? (Just to know)
                $endgroup$
                – Heryon
                Jan 5 at 0:06


















              • $begingroup$
                Clever, thanks. Any way to do this without Taylor series? (Just to know)
                $endgroup$
                – Heryon
                Jan 5 at 0:06
















              $begingroup$
              Clever, thanks. Any way to do this without Taylor series? (Just to know)
              $endgroup$
              – Heryon
              Jan 5 at 0:06




              $begingroup$
              Clever, thanks. Any way to do this without Taylor series? (Just to know)
              $endgroup$
              – Heryon
              Jan 5 at 0:06











              2












              $begingroup$

              Your original function $f(x)=frac{ln(1+x)}{x}$ can be written using the taylor series expansion of $ln(1+x)$ as follows:
              $$ln(1+x)=sum_{k=1}^{infty}frac{(-1)^{k+1}x^{k}}{k}=x-frac{x^2}{2}+frac{x^3}{3}-...$$
              $$thereforefrac{ln(1+x)}{x}=sum_{k=1}^{infty}frac{(-1)^{k+1}x^{k-1}}{k}=sum_{k=0}^{infty}frac{(-1)^{k}x^{k}}{k+1}=1-frac{x}{2}+frac{x^2}{3}-...$$
              $$frac{d}{dx}Bigg(frac{ln(1+x)}{x}Bigg)=frac{d}{dx}bigg(1-frac{x}{2}+frac{x^2}{3}-...bigg)=-frac{1}{2}+frac{2x}{3}-frac{3x^2}{4}+...$$
              So,
              $$lim_{xto0}Bigg(frac{d}{dx}Bigg(frac{ln(1+x)}{x}Bigg)Bigg)=lim_{xto0}bigg(-frac{1}{2}+frac{2x}{3}-frac{3x^2}{4}+...bigg)=fbox{$-frac{1}{2}$}$$






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                Your original function $f(x)=frac{ln(1+x)}{x}$ can be written using the taylor series expansion of $ln(1+x)$ as follows:
                $$ln(1+x)=sum_{k=1}^{infty}frac{(-1)^{k+1}x^{k}}{k}=x-frac{x^2}{2}+frac{x^3}{3}-...$$
                $$thereforefrac{ln(1+x)}{x}=sum_{k=1}^{infty}frac{(-1)^{k+1}x^{k-1}}{k}=sum_{k=0}^{infty}frac{(-1)^{k}x^{k}}{k+1}=1-frac{x}{2}+frac{x^2}{3}-...$$
                $$frac{d}{dx}Bigg(frac{ln(1+x)}{x}Bigg)=frac{d}{dx}bigg(1-frac{x}{2}+frac{x^2}{3}-...bigg)=-frac{1}{2}+frac{2x}{3}-frac{3x^2}{4}+...$$
                So,
                $$lim_{xto0}Bigg(frac{d}{dx}Bigg(frac{ln(1+x)}{x}Bigg)Bigg)=lim_{xto0}bigg(-frac{1}{2}+frac{2x}{3}-frac{3x^2}{4}+...bigg)=fbox{$-frac{1}{2}$}$$






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Your original function $f(x)=frac{ln(1+x)}{x}$ can be written using the taylor series expansion of $ln(1+x)$ as follows:
                  $$ln(1+x)=sum_{k=1}^{infty}frac{(-1)^{k+1}x^{k}}{k}=x-frac{x^2}{2}+frac{x^3}{3}-...$$
                  $$thereforefrac{ln(1+x)}{x}=sum_{k=1}^{infty}frac{(-1)^{k+1}x^{k-1}}{k}=sum_{k=0}^{infty}frac{(-1)^{k}x^{k}}{k+1}=1-frac{x}{2}+frac{x^2}{3}-...$$
                  $$frac{d}{dx}Bigg(frac{ln(1+x)}{x}Bigg)=frac{d}{dx}bigg(1-frac{x}{2}+frac{x^2}{3}-...bigg)=-frac{1}{2}+frac{2x}{3}-frac{3x^2}{4}+...$$
                  So,
                  $$lim_{xto0}Bigg(frac{d}{dx}Bigg(frac{ln(1+x)}{x}Bigg)Bigg)=lim_{xto0}bigg(-frac{1}{2}+frac{2x}{3}-frac{3x^2}{4}+...bigg)=fbox{$-frac{1}{2}$}$$






                  share|cite|improve this answer









                  $endgroup$



                  Your original function $f(x)=frac{ln(1+x)}{x}$ can be written using the taylor series expansion of $ln(1+x)$ as follows:
                  $$ln(1+x)=sum_{k=1}^{infty}frac{(-1)^{k+1}x^{k}}{k}=x-frac{x^2}{2}+frac{x^3}{3}-...$$
                  $$thereforefrac{ln(1+x)}{x}=sum_{k=1}^{infty}frac{(-1)^{k+1}x^{k-1}}{k}=sum_{k=0}^{infty}frac{(-1)^{k}x^{k}}{k+1}=1-frac{x}{2}+frac{x^2}{3}-...$$
                  $$frac{d}{dx}Bigg(frac{ln(1+x)}{x}Bigg)=frac{d}{dx}bigg(1-frac{x}{2}+frac{x^2}{3}-...bigg)=-frac{1}{2}+frac{2x}{3}-frac{3x^2}{4}+...$$
                  So,
                  $$lim_{xto0}Bigg(frac{d}{dx}Bigg(frac{ln(1+x)}{x}Bigg)Bigg)=lim_{xto0}bigg(-frac{1}{2}+frac{2x}{3}-frac{3x^2}{4}+...bigg)=fbox{$-frac{1}{2}$}$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 5 at 0:17









                  Peter ForemanPeter Foreman

                  5,8141317




                  5,8141317























                      1












                      $begingroup$

                      $frac {1}{x(x+1)} - frac {ln(x+1)}{x^2} = frac {d}{dx} frac {ln(x+1)}{x}$



                      At which point I am inclined to use a series



                      $frac {d}{dx} (1 - frac x2 + cdots)$



                      as $x$ goes to $0$ equals $-frac 12$



                      Without using a series, I have.



                      $frac {d}{dx} frac {ln(x+1)}{x}$ evaluated at $x = 0$



                      is $lim_limits{hto 0} frac{ ln(1+h) - 1}{h^2}$






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        Yes I can't use that too, edited the post, but thanks!
                        $endgroup$
                        – Heryon
                        Jan 5 at 0:33
















                      1












                      $begingroup$

                      $frac {1}{x(x+1)} - frac {ln(x+1)}{x^2} = frac {d}{dx} frac {ln(x+1)}{x}$



                      At which point I am inclined to use a series



                      $frac {d}{dx} (1 - frac x2 + cdots)$



                      as $x$ goes to $0$ equals $-frac 12$



                      Without using a series, I have.



                      $frac {d}{dx} frac {ln(x+1)}{x}$ evaluated at $x = 0$



                      is $lim_limits{hto 0} frac{ ln(1+h) - 1}{h^2}$






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        Yes I can't use that too, edited the post, but thanks!
                        $endgroup$
                        – Heryon
                        Jan 5 at 0:33














                      1












                      1








                      1





                      $begingroup$

                      $frac {1}{x(x+1)} - frac {ln(x+1)}{x^2} = frac {d}{dx} frac {ln(x+1)}{x}$



                      At which point I am inclined to use a series



                      $frac {d}{dx} (1 - frac x2 + cdots)$



                      as $x$ goes to $0$ equals $-frac 12$



                      Without using a series, I have.



                      $frac {d}{dx} frac {ln(x+1)}{x}$ evaluated at $x = 0$



                      is $lim_limits{hto 0} frac{ ln(1+h) - 1}{h^2}$






                      share|cite|improve this answer











                      $endgroup$



                      $frac {1}{x(x+1)} - frac {ln(x+1)}{x^2} = frac {d}{dx} frac {ln(x+1)}{x}$



                      At which point I am inclined to use a series



                      $frac {d}{dx} (1 - frac x2 + cdots)$



                      as $x$ goes to $0$ equals $-frac 12$



                      Without using a series, I have.



                      $frac {d}{dx} frac {ln(x+1)}{x}$ evaluated at $x = 0$



                      is $lim_limits{hto 0} frac{ ln(1+h) - 1}{h^2}$







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Jan 5 at 1:30

























                      answered Jan 5 at 0:27









                      Doug MDoug M

                      45.3k31954




                      45.3k31954












                      • $begingroup$
                        Yes I can't use that too, edited the post, but thanks!
                        $endgroup$
                        – Heryon
                        Jan 5 at 0:33


















                      • $begingroup$
                        Yes I can't use that too, edited the post, but thanks!
                        $endgroup$
                        – Heryon
                        Jan 5 at 0:33
















                      $begingroup$
                      Yes I can't use that too, edited the post, but thanks!
                      $endgroup$
                      – Heryon
                      Jan 5 at 0:33




                      $begingroup$
                      Yes I can't use that too, edited the post, but thanks!
                      $endgroup$
                      – Heryon
                      Jan 5 at 0:33











                      0












                      $begingroup$

                      Because $frac{x}{1+x}<ln(1+x)<x$



                      So when $ x rightarrow 0+, f(x) rightarrow 1$



                      When $xrightarrow0-$ is similar






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        Because $frac{x}{1+x}<ln(1+x)<x$



                        So when $ x rightarrow 0+, f(x) rightarrow 1$



                        When $xrightarrow0-$ is similar






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          Because $frac{x}{1+x}<ln(1+x)<x$



                          So when $ x rightarrow 0+, f(x) rightarrow 1$



                          When $xrightarrow0-$ is similar






                          share|cite|improve this answer









                          $endgroup$



                          Because $frac{x}{1+x}<ln(1+x)<x$



                          So when $ x rightarrow 0+, f(x) rightarrow 1$



                          When $xrightarrow0-$ is similar







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 5 at 1:46









                          李子镔李子镔

                          314




                          314






























                              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%2f3062245%2flimlit-of-fracx-x-ln1x-ln1xx3x2-at-0-without-lh%25c3%25b4pital-or-tay%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

                              Probability when a professor distributes a quiz and homework assignment to a class of n students.

                              Image
                              0 0 $begingroup$ Need help with this problem. Suppose our lazy professor collects a quiz and a homework assignment from a class of n students one day, then distributes both the quizzes and the homework assignments back to the class in a random fashion for grading. Each student receives one quiz and one homework assignment to grade. (a) What is the probability that every student receives someone else's quiz to grade, and someone else's homework to grade? (b) What is the probability that no student receives both their own quiz and their own homework assignment to grade? In this case, some students may receive their own quiz, and others may receive their own homework assignment. (c) Compute the limiting probability as n approaches infinity in each case. ...
                              Read more

                              Aardman Animations

                              Image
                              Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
                              Read more

                              Are they similar matrix

                              Image
                              0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
                              Read more