Solve the limit $lim_{xto2}({frac{1}{x(x-2)^2}-frac{1}{x^2-3x+2})}$












0












$begingroup$


$$lim_{xto2}left(frac{1}{x(x-2)^2}-frac{1}{x^2-3x+2}right)$$



What I tried: Got the fraction to the same denominator
$$begin{align}
lim_{xto2}{frac{x^2-3x+2-x(x-2)^2}{x(x-2)^2(x^2-3x+2)}}&=lim_{xto2}{frac{(x-2)(x-1)-x(x-2)^2}{x(x-2)^2(x-2)(x-1)}}\
&=lim_{xto2}{frac{x-1-x^2+2x}{x(x-2)^2(x-1)}}\
&=lim_{xto2}{frac{x-1-x^2+2x}{x(x-2)^2(x-1)}}\
&=lim_{xto2}{frac{x-1-x^2+2x}{x(x-2)^2(x-1)}}\
&=lim_{xto2}{frac{-x^2+3x-1}{x(x-2)^2(x-1)}}\
&={frac{-4+6-1}{0}}
end{align}$$



There is a mistake that I can't find...






calculus limits







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    $$lim_{xto2}left(frac{1}{x(x-2)^2}-frac{1}{x^2-3x+2}right)$$



    What I tried: Got the fraction to the same denominator
    $$begin{align}
    lim_{xto2}{frac{x^2-3x+2-x(x-2)^2}{x(x-2)^2(x^2-3x+2)}}&=lim_{xto2}{frac{(x-2)(x-1)-x(x-2)^2}{x(x-2)^2(x-2)(x-1)}}\
    &=lim_{xto2}{frac{x-1-x^2+2x}{x(x-2)^2(x-1)}}\
    &=lim_{xto2}{frac{x-1-x^2+2x}{x(x-2)^2(x-1)}}\
    &=lim_{xto2}{frac{x-1-x^2+2x}{x(x-2)^2(x-1)}}\
    &=lim_{xto2}{frac{-x^2+3x-1}{x(x-2)^2(x-1)}}\
    &={frac{-4+6-1}{0}}
    end{align}$$



    There is a mistake that I can't find...






    calculus limits







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      $$lim_{xto2}left(frac{1}{x(x-2)^2}-frac{1}{x^2-3x+2}right)$$



      What I tried: Got the fraction to the same denominator
      $$begin{align}
      lim_{xto2}{frac{x^2-3x+2-x(x-2)^2}{x(x-2)^2(x^2-3x+2)}}&=lim_{xto2}{frac{(x-2)(x-1)-x(x-2)^2}{x(x-2)^2(x-2)(x-1)}}\
      &=lim_{xto2}{frac{x-1-x^2+2x}{x(x-2)^2(x-1)}}\
      &=lim_{xto2}{frac{x-1-x^2+2x}{x(x-2)^2(x-1)}}\
      &=lim_{xto2}{frac{x-1-x^2+2x}{x(x-2)^2(x-1)}}\
      &=lim_{xto2}{frac{-x^2+3x-1}{x(x-2)^2(x-1)}}\
      &={frac{-4+6-1}{0}}
      end{align}$$



      There is a mistake that I can't find...






      calculus limits







      share|cite|improve this question











      $endgroup$




      $$lim_{xto2}left(frac{1}{x(x-2)^2}-frac{1}{x^2-3x+2}right)$$



      What I tried: Got the fraction to the same denominator
      $$begin{align}
      lim_{xto2}{frac{x^2-3x+2-x(x-2)^2}{x(x-2)^2(x^2-3x+2)}}&=lim_{xto2}{frac{(x-2)(x-1)-x(x-2)^2}{x(x-2)^2(x-2)(x-1)}}\
      &=lim_{xto2}{frac{x-1-x^2+2x}{x(x-2)^2(x-1)}}\
      &=lim_{xto2}{frac{x-1-x^2+2x}{x(x-2)^2(x-1)}}\
      &=lim_{xto2}{frac{x-1-x^2+2x}{x(x-2)^2(x-1)}}\
      &=lim_{xto2}{frac{-x^2+3x-1}{x(x-2)^2(x-1)}}\
      &={frac{-4+6-1}{0}}
      end{align}$$



      There is a mistake that I can't find...






      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 Dec 8 '18 at 14:50









      mrtaurho

      4,23421234




      4,23421234










      asked Dec 8 '18 at 14:33









      Bili DebiliBili Debili

      1428




      1428






















          3 Answers
          3






          active

          oldest

          votes


















          3












          $begingroup$

          The numerator tend to a positive number, the denominator goes to $0$ from the positive directione. The sequence actually goes to infinity.



          enter image description here






          share|cite|improve this answer









          $endgroup$





















            2












            $begingroup$

            Hint: Use that $$x^2-3x+2-x(x-2)^2=-(x-2)(x^2-3x+1)$$ and $$x^2-3x+2=(x-1)(x-2)$$






            share|cite|improve this answer









            $endgroup$





















              1












              $begingroup$

              By $y=x-2 to 0$ we have that



              $$lim_{xto 2} left({frac{1}{x(x-2)^2}-frac{1}{x^2-3x+2}}right)=lim_{yto 0} left({frac{1}{y^2(y+2)}-frac{1}{y(y+1)}}right)$$



              and



              $${frac{1}{y^2(y+2)}-frac{1}{y(y+1)}}={frac{y+1-y^2-2y}{y^2(y+1)(y+2)}}=frac1ycdot{frac{-y^2-y+1}{(y+1)(y+2)}}to infty$$






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                @user376343 mmmmhhhhh yes I thing there is something wrong with that! I fix...thanks
                $endgroup$
                – gimusi
                Dec 8 '18 at 18:58











              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%2f3031174%2fsolve-the-limit-lim-x-to2-frac1xx-22-frac1x2-3x2%23new-answer', 'question_page');
              }
              );

              Post as a guest















              Required, but never shown

























              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              3












              $begingroup$

              The numerator tend to a positive number, the denominator goes to $0$ from the positive directione. The sequence actually goes to infinity.



              enter image description here






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                The numerator tend to a positive number, the denominator goes to $0$ from the positive directione. The sequence actually goes to infinity.



                enter image description here






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  The numerator tend to a positive number, the denominator goes to $0$ from the positive directione. The sequence actually goes to infinity.



                  enter image description here






                  share|cite|improve this answer









                  $endgroup$



                  The numerator tend to a positive number, the denominator goes to $0$ from the positive directione. The sequence actually goes to infinity.



                  enter image description here







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 8 '18 at 14:40









                  Siong Thye GohSiong Thye Goh

                  101k1466117




                  101k1466117























                      2












                      $begingroup$

                      Hint: Use that $$x^2-3x+2-x(x-2)^2=-(x-2)(x^2-3x+1)$$ and $$x^2-3x+2=(x-1)(x-2)$$






                      share|cite|improve this answer









                      $endgroup$


















                        2












                        $begingroup$

                        Hint: Use that $$x^2-3x+2-x(x-2)^2=-(x-2)(x^2-3x+1)$$ and $$x^2-3x+2=(x-1)(x-2)$$






                        share|cite|improve this answer









                        $endgroup$
















                          2












                          2








                          2





                          $begingroup$

                          Hint: Use that $$x^2-3x+2-x(x-2)^2=-(x-2)(x^2-3x+1)$$ and $$x^2-3x+2=(x-1)(x-2)$$






                          share|cite|improve this answer









                          $endgroup$



                          Hint: Use that $$x^2-3x+2-x(x-2)^2=-(x-2)(x^2-3x+1)$$ and $$x^2-3x+2=(x-1)(x-2)$$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 8 '18 at 14:39









                          Dr. Sonnhard GraubnerDr. Sonnhard Graubner

                          74.6k42865




                          74.6k42865























                              1












                              $begingroup$

                              By $y=x-2 to 0$ we have that



                              $$lim_{xto 2} left({frac{1}{x(x-2)^2}-frac{1}{x^2-3x+2}}right)=lim_{yto 0} left({frac{1}{y^2(y+2)}-frac{1}{y(y+1)}}right)$$



                              and



                              $${frac{1}{y^2(y+2)}-frac{1}{y(y+1)}}={frac{y+1-y^2-2y}{y^2(y+1)(y+2)}}=frac1ycdot{frac{-y^2-y+1}{(y+1)(y+2)}}to infty$$






                              share|cite|improve this answer











                              $endgroup$













                              • $begingroup$
                                @user376343 mmmmhhhhh yes I thing there is something wrong with that! I fix...thanks
                                $endgroup$
                                – gimusi
                                Dec 8 '18 at 18:58
















                              1












                              $begingroup$

                              By $y=x-2 to 0$ we have that



                              $$lim_{xto 2} left({frac{1}{x(x-2)^2}-frac{1}{x^2-3x+2}}right)=lim_{yto 0} left({frac{1}{y^2(y+2)}-frac{1}{y(y+1)}}right)$$



                              and



                              $${frac{1}{y^2(y+2)}-frac{1}{y(y+1)}}={frac{y+1-y^2-2y}{y^2(y+1)(y+2)}}=frac1ycdot{frac{-y^2-y+1}{(y+1)(y+2)}}to infty$$






                              share|cite|improve this answer











                              $endgroup$













                              • $begingroup$
                                @user376343 mmmmhhhhh yes I thing there is something wrong with that! I fix...thanks
                                $endgroup$
                                – gimusi
                                Dec 8 '18 at 18:58














                              1












                              1








                              1





                              $begingroup$

                              By $y=x-2 to 0$ we have that



                              $$lim_{xto 2} left({frac{1}{x(x-2)^2}-frac{1}{x^2-3x+2}}right)=lim_{yto 0} left({frac{1}{y^2(y+2)}-frac{1}{y(y+1)}}right)$$



                              and



                              $${frac{1}{y^2(y+2)}-frac{1}{y(y+1)}}={frac{y+1-y^2-2y}{y^2(y+1)(y+2)}}=frac1ycdot{frac{-y^2-y+1}{(y+1)(y+2)}}to infty$$






                              share|cite|improve this answer











                              $endgroup$



                              By $y=x-2 to 0$ we have that



                              $$lim_{xto 2} left({frac{1}{x(x-2)^2}-frac{1}{x^2-3x+2}}right)=lim_{yto 0} left({frac{1}{y^2(y+2)}-frac{1}{y(y+1)}}right)$$



                              and



                              $${frac{1}{y^2(y+2)}-frac{1}{y(y+1)}}={frac{y+1-y^2-2y}{y^2(y+1)(y+2)}}=frac1ycdot{frac{-y^2-y+1}{(y+1)(y+2)}}to infty$$







                              share|cite|improve this answer














                              share|cite|improve this answer



                              share|cite|improve this answer








                              edited Dec 8 '18 at 18:59

























                              answered Dec 8 '18 at 14:41









                              gimusigimusi

                              92.8k84494




                              92.8k84494












                              • $begingroup$
                                @user376343 mmmmhhhhh yes I thing there is something wrong with that! I fix...thanks
                                $endgroup$
                                – gimusi
                                Dec 8 '18 at 18:58


















                              • $begingroup$
                                @user376343 mmmmhhhhh yes I thing there is something wrong with that! I fix...thanks
                                $endgroup$
                                – gimusi
                                Dec 8 '18 at 18:58
















                              $begingroup$
                              @user376343 mmmmhhhhh yes I thing there is something wrong with that! I fix...thanks
                              $endgroup$
                              – gimusi
                              Dec 8 '18 at 18:58




                              $begingroup$
                              @user376343 mmmmhhhhh yes I thing there is something wrong with that! I fix...thanks
                              $endgroup$
                              – gimusi
                              Dec 8 '18 at 18:58


















                              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%2f3031174%2fsolve-the-limit-lim-x-to2-frac1xx-22-frac1x2-3x2%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

                              Quarter-circle Tiles

                              Image
                              up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2
                              Read more

                              build a pushdown automaton that recognizes the reverse language of a given pushdown automaton?

                              Image
                              0 I think it's similiar to NFAs. I replace the finite states of the given automaton for start-states for my new automaton. I do it with $epsilon$ -transition from the start state to the actual finite states of the given automaton. The start state of the given automaton will be my accepting(finite) state of my new one. All transitions will be reversed of course to recognize the reversed word. But how do I have to change the stack of the pushdown automaton? That's tricky. formal-languages automata context-free-grammar share | cite | improve this question edited Nov 29 '18 at 14:20 joee
                              Read more

                              Mont Emei

                              Image
                              Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog
                              Read more