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











up vote
2
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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







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    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





















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                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






























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