solve $lim_{xtoinfty}{sqrt{x^4+x^2}+sqrt{x^2+5x}-x^2-x}$











up vote
2
down vote

favorite












I tried solve this limit.



$lim_{xtoinfty}{sqrt{x^4+x^2}+sqrt{x^2+5x}-x^2-x}$



I tried multiply by $sqrt{x^4+x^2}-sqrt{x^2+5x}$, and apply L'Hospital. but this led to alot of work.. and this question seems to had a very easy and fast way to do. I know the answer is 3.



thanks any help.






calculus







share|cite|improve this question


























    up vote
    2
    down vote

    favorite












    I tried solve this limit.



    $lim_{xtoinfty}{sqrt{x^4+x^2}+sqrt{x^2+5x}-x^2-x}$



    I tried multiply by $sqrt{x^4+x^2}-sqrt{x^2+5x}$, and apply L'Hospital. but this led to alot of work.. and this question seems to had a very easy and fast way to do. I know the answer is 3.



    thanks any help.






    calculus







    share|cite|improve this question
























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      I tried solve this limit.



      $lim_{xtoinfty}{sqrt{x^4+x^2}+sqrt{x^2+5x}-x^2-x}$



      I tried multiply by $sqrt{x^4+x^2}-sqrt{x^2+5x}$, and apply L'Hospital. but this led to alot of work.. and this question seems to had a very easy and fast way to do. I know the answer is 3.



      thanks any help.






      calculus







      share|cite|improve this question













      I tried solve this limit.



      $lim_{xtoinfty}{sqrt{x^4+x^2}+sqrt{x^2+5x}-x^2-x}$



      I tried multiply by $sqrt{x^4+x^2}-sqrt{x^2+5x}$, and apply L'Hospital. but this led to alot of work.. and this question seems to had a very easy and fast way to do. I know the answer is 3.



      thanks any help.






      calculus




      calculus






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 22 at 16:58









      Juliana Neves

      152




      152






















          4 Answers
          4






          active

          oldest

          votes

















          up vote
          2
          down vote



          accepted










          Hint:



          Compute separately
          $$lim_{xtoinfty}sqrt{x^4+x^2}-x^2$$
          and
          $$lim_{xtoinfty}sqrt{x^2+5x}-x$$






          share|cite|improve this answer




























            up vote
            1
            down vote













            Continue with ajotatxe's hint:



            $$lim_{x to infty} sqrt{x^4 + x^2} - x^2$$



            $$lim_{x to infty} x^2 (sqrt{1 + frac{1}{x^2}} - 1)$$



            $$lim_{x to infty} frac{x^2 (sqrt{1 + frac{1}{x^2}} - 1)(sqrt{1 + frac{1}{x^2}} + 1)}{sqrt{1 + frac{1}{x^2}} + 1} $$



            $$lim_{x to infty} frac{x^2 frac{1}{x^2}}{sqrt{1 + frac{1}{x^2}} + 1} $$



            $$lim_{x to infty} frac{1}{sqrt{1 + frac{1}{x^2}} + 1} $$






            share|cite|improve this answer




























              up vote
              0
              down vote













              begin{align}
              lim_{xtoinfty}{sqrt{x^4+x^2}+sqrt{x^2+5x}-x^2-x}
              &= lim_{xtoinfty}sqrt{left(x^2+dfrac12right)^2-dfrac14} \
              &+sqrt{left(x+frac52right)^2-frac{25}{4}} -x^2-x \
              &= lim_{xtoinfty}{left(x^2+dfrac12right)+left|x+frac52right|-x^2-x} \
              &= lim_{xto+infty}{dfrac12+x+frac52-x} \
              &= color{blue}{3}
              end{align}






              share|cite|improve this answer























              • This is as wrong as can be, since this suggests, for example, that $$sqrt{x^4+x^2}-x^2to0$$ which is obviously not the case. @upvoter Why the upvote?
                – Did
                Nov 25 at 9:41




















              up vote
              -1
              down vote













              HINT



              We have that by binomial expansion



              $${sqrt{x^4+x^2}+sqrt{x^2+5x}-x^2-x}=x^2(1+1/x^2)^frac12+x(1+5/x)^frac12-x^2-x=$$



              $$=x^2+frac12+x+frac52-x^2-x+oleft(frac1xright)$$






              share|cite|improve this answer





















                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',
                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%2f3009361%2fsolve-lim-x-to-infty-sqrtx4x2-sqrtx25x-x2-x%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








                up vote
                2
                down vote



                accepted










                Hint:



                Compute separately
                $$lim_{xtoinfty}sqrt{x^4+x^2}-x^2$$
                and
                $$lim_{xtoinfty}sqrt{x^2+5x}-x$$






                share|cite|improve this answer

























                  up vote
                  2
                  down vote



                  accepted










                  Hint:



                  Compute separately
                  $$lim_{xtoinfty}sqrt{x^4+x^2}-x^2$$
                  and
                  $$lim_{xtoinfty}sqrt{x^2+5x}-x$$






                  share|cite|improve this answer























                    up vote
                    2
                    down vote



                    accepted







                    up vote
                    2
                    down vote



                    accepted






                    Hint:



                    Compute separately
                    $$lim_{xtoinfty}sqrt{x^4+x^2}-x^2$$
                    and
                    $$lim_{xtoinfty}sqrt{x^2+5x}-x$$






                    share|cite|improve this answer












                    Hint:



                    Compute separately
                    $$lim_{xtoinfty}sqrt{x^4+x^2}-x^2$$
                    and
                    $$lim_{xtoinfty}sqrt{x^2+5x}-x$$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Nov 22 at 17:01









                    ajotatxe

                    53k23890




                    53k23890






















                        up vote
                        1
                        down vote













                        Continue with ajotatxe's hint:



                        $$lim_{x to infty} sqrt{x^4 + x^2} - x^2$$



                        $$lim_{x to infty} x^2 (sqrt{1 + frac{1}{x^2}} - 1)$$



                        $$lim_{x to infty} frac{x^2 (sqrt{1 + frac{1}{x^2}} - 1)(sqrt{1 + frac{1}{x^2}} + 1)}{sqrt{1 + frac{1}{x^2}} + 1} $$



                        $$lim_{x to infty} frac{x^2 frac{1}{x^2}}{sqrt{1 + frac{1}{x^2}} + 1} $$



                        $$lim_{x to infty} frac{1}{sqrt{1 + frac{1}{x^2}} + 1} $$






                        share|cite|improve this answer

























                          up vote
                          1
                          down vote













                          Continue with ajotatxe's hint:



                          $$lim_{x to infty} sqrt{x^4 + x^2} - x^2$$



                          $$lim_{x to infty} x^2 (sqrt{1 + frac{1}{x^2}} - 1)$$



                          $$lim_{x to infty} frac{x^2 (sqrt{1 + frac{1}{x^2}} - 1)(sqrt{1 + frac{1}{x^2}} + 1)}{sqrt{1 + frac{1}{x^2}} + 1} $$



                          $$lim_{x to infty} frac{x^2 frac{1}{x^2}}{sqrt{1 + frac{1}{x^2}} + 1} $$



                          $$lim_{x to infty} frac{1}{sqrt{1 + frac{1}{x^2}} + 1} $$






                          share|cite|improve this answer























                            up vote
                            1
                            down vote










                            up vote
                            1
                            down vote









                            Continue with ajotatxe's hint:



                            $$lim_{x to infty} sqrt{x^4 + x^2} - x^2$$



                            $$lim_{x to infty} x^2 (sqrt{1 + frac{1}{x^2}} - 1)$$



                            $$lim_{x to infty} frac{x^2 (sqrt{1 + frac{1}{x^2}} - 1)(sqrt{1 + frac{1}{x^2}} + 1)}{sqrt{1 + frac{1}{x^2}} + 1} $$



                            $$lim_{x to infty} frac{x^2 frac{1}{x^2}}{sqrt{1 + frac{1}{x^2}} + 1} $$



                            $$lim_{x to infty} frac{1}{sqrt{1 + frac{1}{x^2}} + 1} $$






                            share|cite|improve this answer












                            Continue with ajotatxe's hint:



                            $$lim_{x to infty} sqrt{x^4 + x^2} - x^2$$



                            $$lim_{x to infty} x^2 (sqrt{1 + frac{1}{x^2}} - 1)$$



                            $$lim_{x to infty} frac{x^2 (sqrt{1 + frac{1}{x^2}} - 1)(sqrt{1 + frac{1}{x^2}} + 1)}{sqrt{1 + frac{1}{x^2}} + 1} $$



                            $$lim_{x to infty} frac{x^2 frac{1}{x^2}}{sqrt{1 + frac{1}{x^2}} + 1} $$



                            $$lim_{x to infty} frac{1}{sqrt{1 + frac{1}{x^2}} + 1} $$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Nov 22 at 17:26









                            chhhh

                            242




                            242






















                                up vote
                                0
                                down vote













                                begin{align}
                                lim_{xtoinfty}{sqrt{x^4+x^2}+sqrt{x^2+5x}-x^2-x}
                                &= lim_{xtoinfty}sqrt{left(x^2+dfrac12right)^2-dfrac14} \
                                &+sqrt{left(x+frac52right)^2-frac{25}{4}} -x^2-x \
                                &= lim_{xtoinfty}{left(x^2+dfrac12right)+left|x+frac52right|-x^2-x} \
                                &= lim_{xto+infty}{dfrac12+x+frac52-x} \
                                &= color{blue}{3}
                                end{align}






                                share|cite|improve this answer























                                • This is as wrong as can be, since this suggests, for example, that $$sqrt{x^4+x^2}-x^2to0$$ which is obviously not the case. @upvoter Why the upvote?
                                  – Did
                                  Nov 25 at 9:41

















                                up vote
                                0
                                down vote













                                begin{align}
                                lim_{xtoinfty}{sqrt{x^4+x^2}+sqrt{x^2+5x}-x^2-x}
                                &= lim_{xtoinfty}sqrt{left(x^2+dfrac12right)^2-dfrac14} \
                                &+sqrt{left(x+frac52right)^2-frac{25}{4}} -x^2-x \
                                &= lim_{xtoinfty}{left(x^2+dfrac12right)+left|x+frac52right|-x^2-x} \
                                &= lim_{xto+infty}{dfrac12+x+frac52-x} \
                                &= color{blue}{3}
                                end{align}






                                share|cite|improve this answer























                                • This is as wrong as can be, since this suggests, for example, that $$sqrt{x^4+x^2}-x^2to0$$ which is obviously not the case. @upvoter Why the upvote?
                                  – Did
                                  Nov 25 at 9:41















                                up vote
                                0
                                down vote










                                up vote
                                0
                                down vote









                                begin{align}
                                lim_{xtoinfty}{sqrt{x^4+x^2}+sqrt{x^2+5x}-x^2-x}
                                &= lim_{xtoinfty}sqrt{left(x^2+dfrac12right)^2-dfrac14} \
                                &+sqrt{left(x+frac52right)^2-frac{25}{4}} -x^2-x \
                                &= lim_{xtoinfty}{left(x^2+dfrac12right)+left|x+frac52right|-x^2-x} \
                                &= lim_{xto+infty}{dfrac12+x+frac52-x} \
                                &= color{blue}{3}
                                end{align}






                                share|cite|improve this answer














                                begin{align}
                                lim_{xtoinfty}{sqrt{x^4+x^2}+sqrt{x^2+5x}-x^2-x}
                                &= lim_{xtoinfty}sqrt{left(x^2+dfrac12right)^2-dfrac14} \
                                &+sqrt{left(x+frac52right)^2-frac{25}{4}} -x^2-x \
                                &= lim_{xtoinfty}{left(x^2+dfrac12right)+left|x+frac52right|-x^2-x} \
                                &= lim_{xto+infty}{dfrac12+x+frac52-x} \
                                &= color{blue}{3}
                                end{align}







                                share|cite|improve this answer














                                share|cite|improve this answer



                                share|cite|improve this answer








                                edited Nov 25 at 11:22

























                                answered Nov 22 at 17:07









                                Nosrati

                                26.3k62353




                                26.3k62353












                                • This is as wrong as can be, since this suggests, for example, that $$sqrt{x^4+x^2}-x^2to0$$ which is obviously not the case. @upvoter Why the upvote?
                                  – Did
                                  Nov 25 at 9:41




















                                • This is as wrong as can be, since this suggests, for example, that $$sqrt{x^4+x^2}-x^2to0$$ which is obviously not the case. @upvoter Why the upvote?
                                  – Did
                                  Nov 25 at 9:41


















                                This is as wrong as can be, since this suggests, for example, that $$sqrt{x^4+x^2}-x^2to0$$ which is obviously not the case. @upvoter Why the upvote?
                                – Did
                                Nov 25 at 9:41






                                This is as wrong as can be, since this suggests, for example, that $$sqrt{x^4+x^2}-x^2to0$$ which is obviously not the case. @upvoter Why the upvote?
                                – Did
                                Nov 25 at 9:41












                                up vote
                                -1
                                down vote













                                HINT



                                We have that by binomial expansion



                                $${sqrt{x^4+x^2}+sqrt{x^2+5x}-x^2-x}=x^2(1+1/x^2)^frac12+x(1+5/x)^frac12-x^2-x=$$



                                $$=x^2+frac12+x+frac52-x^2-x+oleft(frac1xright)$$






                                share|cite|improve this answer

























                                  up vote
                                  -1
                                  down vote













                                  HINT



                                  We have that by binomial expansion



                                  $${sqrt{x^4+x^2}+sqrt{x^2+5x}-x^2-x}=x^2(1+1/x^2)^frac12+x(1+5/x)^frac12-x^2-x=$$



                                  $$=x^2+frac12+x+frac52-x^2-x+oleft(frac1xright)$$






                                  share|cite|improve this answer























                                    up vote
                                    -1
                                    down vote










                                    up vote
                                    -1
                                    down vote









                                    HINT



                                    We have that by binomial expansion



                                    $${sqrt{x^4+x^2}+sqrt{x^2+5x}-x^2-x}=x^2(1+1/x^2)^frac12+x(1+5/x)^frac12-x^2-x=$$



                                    $$=x^2+frac12+x+frac52-x^2-x+oleft(frac1xright)$$






                                    share|cite|improve this answer












                                    HINT



                                    We have that by binomial expansion



                                    $${sqrt{x^4+x^2}+sqrt{x^2+5x}-x^2-x}=x^2(1+1/x^2)^frac12+x(1+5/x)^frac12-x^2-x=$$



                                    $$=x^2+frac12+x+frac52-x^2-x+oleft(frac1xright)$$







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Nov 22 at 17:16









                                    gimusi

                                    92.7k94495




                                    92.7k94495






























                                        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.





                                        Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                                        Please pay close attention to the following guidance:


                                        • 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.


                                        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%2f3009361%2fsolve-lim-x-to-infty-sqrtx4x2-sqrtx25x-x2-x%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