Evaluation of the limit $displaystylelim_{xtoinfty}x-sqrt{(x-a)(x-b)}$












0












$begingroup$


I'm trying to evaluate the limit $displaystylelim_{xtoinfty}x-sqrt{(x-a)(x-b)}$




Since there is a square root of a product namely $sqrt{(x-a)(x-b)}$ I
think it's useful to use $text{GM}letext{AM}$ for $(x-a)$ & $(x-b)$:
$sqrt{(x-a)(x-b)}ledfrac{(x-a)+(x-b)}{2}=x-dfrac{a+b}{2}tag*{}$



To see whether the function $f(x)=x-sqrt{(x-a)(x-b)}$ is increasing
or decreasing,



begin{align}f'(x)&=1-dfrac{2x-(a+b)}{2sqrt{(x-a)(x-b)}}\&=1-dfrac{dfrac{x-(a+b)}2}{sqrt{(x-a)(x-b)}}\&=1-dfrac{text{AM}}{GM}\&le0
becausetext{ GM}letext{AM}end{align}



hence $f(x)=x-sqrt{(x-a)(x-b)}$ is decreasing. Then in order for the
existence of the limit, $f(x)$ must be bounded below.



Here I again use the same inequality:



begin{align}\&sqrt{(x-a)(x-b)}ledfrac{(x-a)+(x-b)}{2}=x-dfrac{a+b}{2}\&implies
f(x)=x-sqrt{(x-a)(x-b)}gedfrac{a+b}{2}end{align}



If I'm right, then the limit exists and
$displaystylelim_{xtoinfty}x-sqrt{(x-a)(x-b)}=dfrac{a+b}2$




Another query is that in the inequality $sqrt{(x-a)(x-b)}ledfrac{(x-a)+(x-b)}{2}$ the equality holds iff $x-a=x-b$- how to solve it for $x$? I want to know where $f(x)$ takes its minimum.






real-analysis calculus limits proof-verification







share|cite|improve this question











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  • 1




    $begingroup$
    It's worth noting that your AM/GM argument, while very nice and well presented, only proves that the limit exists and is greater than or equal to $(a+b)/2$. Something more is required to show that it's actually equal to $(a+b)/2$.
    $endgroup$
    – Barry Cipra
    Jan 5 at 14:24












  • $begingroup$
    @ Barry Cipra can you please suggest edits to my answer?
    $endgroup$
    – Arjun Banerjee
    Jan 5 at 14:28










  • $begingroup$
    I was just about to add that I'd actually like to see your argument completed, because nothing obvious comes to (my) mind that doesn't effectively replace the entire proof with the approach given in Robert Z's answer. It's worth thinking about!
    $endgroup$
    – Barry Cipra
    Jan 5 at 14:30
















0












$begingroup$


I'm trying to evaluate the limit $displaystylelim_{xtoinfty}x-sqrt{(x-a)(x-b)}$




Since there is a square root of a product namely $sqrt{(x-a)(x-b)}$ I
think it's useful to use $text{GM}letext{AM}$ for $(x-a)$ & $(x-b)$:
$sqrt{(x-a)(x-b)}ledfrac{(x-a)+(x-b)}{2}=x-dfrac{a+b}{2}tag*{}$



To see whether the function $f(x)=x-sqrt{(x-a)(x-b)}$ is increasing
or decreasing,



begin{align}f'(x)&=1-dfrac{2x-(a+b)}{2sqrt{(x-a)(x-b)}}\&=1-dfrac{dfrac{x-(a+b)}2}{sqrt{(x-a)(x-b)}}\&=1-dfrac{text{AM}}{GM}\&le0
becausetext{ GM}letext{AM}end{align}



hence $f(x)=x-sqrt{(x-a)(x-b)}$ is decreasing. Then in order for the
existence of the limit, $f(x)$ must be bounded below.



Here I again use the same inequality:



begin{align}\&sqrt{(x-a)(x-b)}ledfrac{(x-a)+(x-b)}{2}=x-dfrac{a+b}{2}\&implies
f(x)=x-sqrt{(x-a)(x-b)}gedfrac{a+b}{2}end{align}



If I'm right, then the limit exists and
$displaystylelim_{xtoinfty}x-sqrt{(x-a)(x-b)}=dfrac{a+b}2$




Another query is that in the inequality $sqrt{(x-a)(x-b)}ledfrac{(x-a)+(x-b)}{2}$ the equality holds iff $x-a=x-b$- how to solve it for $x$? I want to know where $f(x)$ takes its minimum.






real-analysis calculus limits proof-verification







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    It's worth noting that your AM/GM argument, while very nice and well presented, only proves that the limit exists and is greater than or equal to $(a+b)/2$. Something more is required to show that it's actually equal to $(a+b)/2$.
    $endgroup$
    – Barry Cipra
    Jan 5 at 14:24












  • $begingroup$
    @ Barry Cipra can you please suggest edits to my answer?
    $endgroup$
    – Arjun Banerjee
    Jan 5 at 14:28










  • $begingroup$
    I was just about to add that I'd actually like to see your argument completed, because nothing obvious comes to (my) mind that doesn't effectively replace the entire proof with the approach given in Robert Z's answer. It's worth thinking about!
    $endgroup$
    – Barry Cipra
    Jan 5 at 14:30














0












0








0


2



$begingroup$


I'm trying to evaluate the limit $displaystylelim_{xtoinfty}x-sqrt{(x-a)(x-b)}$




Since there is a square root of a product namely $sqrt{(x-a)(x-b)}$ I
think it's useful to use $text{GM}letext{AM}$ for $(x-a)$ & $(x-b)$:
$sqrt{(x-a)(x-b)}ledfrac{(x-a)+(x-b)}{2}=x-dfrac{a+b}{2}tag*{}$



To see whether the function $f(x)=x-sqrt{(x-a)(x-b)}$ is increasing
or decreasing,



begin{align}f'(x)&=1-dfrac{2x-(a+b)}{2sqrt{(x-a)(x-b)}}\&=1-dfrac{dfrac{x-(a+b)}2}{sqrt{(x-a)(x-b)}}\&=1-dfrac{text{AM}}{GM}\&le0
becausetext{ GM}letext{AM}end{align}



hence $f(x)=x-sqrt{(x-a)(x-b)}$ is decreasing. Then in order for the
existence of the limit, $f(x)$ must be bounded below.



Here I again use the same inequality:



begin{align}\&sqrt{(x-a)(x-b)}ledfrac{(x-a)+(x-b)}{2}=x-dfrac{a+b}{2}\&implies
f(x)=x-sqrt{(x-a)(x-b)}gedfrac{a+b}{2}end{align}



If I'm right, then the limit exists and
$displaystylelim_{xtoinfty}x-sqrt{(x-a)(x-b)}=dfrac{a+b}2$




Another query is that in the inequality $sqrt{(x-a)(x-b)}ledfrac{(x-a)+(x-b)}{2}$ the equality holds iff $x-a=x-b$- how to solve it for $x$? I want to know where $f(x)$ takes its minimum.






real-analysis calculus limits proof-verification







share|cite|improve this question











$endgroup$




I'm trying to evaluate the limit $displaystylelim_{xtoinfty}x-sqrt{(x-a)(x-b)}$




Since there is a square root of a product namely $sqrt{(x-a)(x-b)}$ I
think it's useful to use $text{GM}letext{AM}$ for $(x-a)$ & $(x-b)$:
$sqrt{(x-a)(x-b)}ledfrac{(x-a)+(x-b)}{2}=x-dfrac{a+b}{2}tag*{}$



To see whether the function $f(x)=x-sqrt{(x-a)(x-b)}$ is increasing
or decreasing,



begin{align}f'(x)&=1-dfrac{2x-(a+b)}{2sqrt{(x-a)(x-b)}}\&=1-dfrac{dfrac{x-(a+b)}2}{sqrt{(x-a)(x-b)}}\&=1-dfrac{text{AM}}{GM}\&le0
becausetext{ GM}letext{AM}end{align}



hence $f(x)=x-sqrt{(x-a)(x-b)}$ is decreasing. Then in order for the
existence of the limit, $f(x)$ must be bounded below.



Here I again use the same inequality:



begin{align}\&sqrt{(x-a)(x-b)}ledfrac{(x-a)+(x-b)}{2}=x-dfrac{a+b}{2}\&implies
f(x)=x-sqrt{(x-a)(x-b)}gedfrac{a+b}{2}end{align}



If I'm right, then the limit exists and
$displaystylelim_{xtoinfty}x-sqrt{(x-a)(x-b)}=dfrac{a+b}2$




Another query is that in the inequality $sqrt{(x-a)(x-b)}ledfrac{(x-a)+(x-b)}{2}$ the equality holds iff $x-a=x-b$- how to solve it for $x$? I want to know where $f(x)$ takes its minimum.






real-analysis calculus limits proof-verification




real-analysis calculus limits proof-verification






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 5 at 13:41







Arjun Banerjee

















asked Jan 5 at 13:38









Arjun BanerjeeArjun Banerjee

54210




54210








  • 1




    $begingroup$
    It's worth noting that your AM/GM argument, while very nice and well presented, only proves that the limit exists and is greater than or equal to $(a+b)/2$. Something more is required to show that it's actually equal to $(a+b)/2$.
    $endgroup$
    – Barry Cipra
    Jan 5 at 14:24












  • $begingroup$
    @ Barry Cipra can you please suggest edits to my answer?
    $endgroup$
    – Arjun Banerjee
    Jan 5 at 14:28










  • $begingroup$
    I was just about to add that I'd actually like to see your argument completed, because nothing obvious comes to (my) mind that doesn't effectively replace the entire proof with the approach given in Robert Z's answer. It's worth thinking about!
    $endgroup$
    – Barry Cipra
    Jan 5 at 14:30














  • 1




    $begingroup$
    It's worth noting that your AM/GM argument, while very nice and well presented, only proves that the limit exists and is greater than or equal to $(a+b)/2$. Something more is required to show that it's actually equal to $(a+b)/2$.
    $endgroup$
    – Barry Cipra
    Jan 5 at 14:24












  • $begingroup$
    @ Barry Cipra can you please suggest edits to my answer?
    $endgroup$
    – Arjun Banerjee
    Jan 5 at 14:28










  • $begingroup$
    I was just about to add that I'd actually like to see your argument completed, because nothing obvious comes to (my) mind that doesn't effectively replace the entire proof with the approach given in Robert Z's answer. It's worth thinking about!
    $endgroup$
    – Barry Cipra
    Jan 5 at 14:30








1




1




$begingroup$
It's worth noting that your AM/GM argument, while very nice and well presented, only proves that the limit exists and is greater than or equal to $(a+b)/2$. Something more is required to show that it's actually equal to $(a+b)/2$.
$endgroup$
– Barry Cipra
Jan 5 at 14:24






$begingroup$
It's worth noting that your AM/GM argument, while very nice and well presented, only proves that the limit exists and is greater than or equal to $(a+b)/2$. Something more is required to show that it's actually equal to $(a+b)/2$.
$endgroup$
– Barry Cipra
Jan 5 at 14:24














$begingroup$
@ Barry Cipra can you please suggest edits to my answer?
$endgroup$
– Arjun Banerjee
Jan 5 at 14:28




$begingroup$
@ Barry Cipra can you please suggest edits to my answer?
$endgroup$
– Arjun Banerjee
Jan 5 at 14:28












$begingroup$
I was just about to add that I'd actually like to see your argument completed, because nothing obvious comes to (my) mind that doesn't effectively replace the entire proof with the approach given in Robert Z's answer. It's worth thinking about!
$endgroup$
– Barry Cipra
Jan 5 at 14:30




$begingroup$
I was just about to add that I'd actually like to see your argument completed, because nothing obvious comes to (my) mind that doesn't effectively replace the entire proof with the approach given in Robert Z's answer. It's worth thinking about!
$endgroup$
– Barry Cipra
Jan 5 at 14:30










3 Answers
3






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oldest

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4












$begingroup$

You are correct, but the limit is easier to evaluate if you consider
$$x-sqrt{(x-a)(x-b)}=frac{x^2-(x-a)(x-b)}{x+sqrt{(x-a)(x-b)}}=
frac{(a+b)-ab/x}{1+sqrt{(1-a/x)(1-b/x)}}$$
for $x> max(a,b,0)$.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Alternatively, do the change of variable $x = 1/t$
    $$lim_{xto+infty}x - sqrt{(x-a)(x-b)} = lim_{tto 0^+}frac{1 - sqrt{1 - (a + b)t + abt^2}}t$$
    and apply Taylor:
    $$sqrt{1 - (a + b)t + abt^2} = 1 - frac{a + b}2 t + cdots$$






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      Hint:



      Evaluation of the limit $displaystylelim_{xtoinfty}x-sqrt{(x-a)(x-b)}$,



      Rationalizing the numerator,
      $$lim_{tto0^+}dfrac{1-sqrt{1-(a+b)t+abt^2}}t=lim_{tto0^+}dfrac{1-(1-(a+b)t+abt^2)}tcdotlim_{tto0^+}dfrac1{1+sqrt{1-(a+b)t+abt^2}}$$



      As $tto0,tne0,dfrac{1-(1-(a+b)t+abt^2)}t=(a+b)-abt$






      share|cite|improve this answer









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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

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        4












        $begingroup$

        You are correct, but the limit is easier to evaluate if you consider
        $$x-sqrt{(x-a)(x-b)}=frac{x^2-(x-a)(x-b)}{x+sqrt{(x-a)(x-b)}}=
        frac{(a+b)-ab/x}{1+sqrt{(1-a/x)(1-b/x)}}$$
        for $x> max(a,b,0)$.






        share|cite|improve this answer









        $endgroup$


















          4












          $begingroup$

          You are correct, but the limit is easier to evaluate if you consider
          $$x-sqrt{(x-a)(x-b)}=frac{x^2-(x-a)(x-b)}{x+sqrt{(x-a)(x-b)}}=
          frac{(a+b)-ab/x}{1+sqrt{(1-a/x)(1-b/x)}}$$
          for $x> max(a,b,0)$.






          share|cite|improve this answer









          $endgroup$
















            4












            4








            4





            $begingroup$

            You are correct, but the limit is easier to evaluate if you consider
            $$x-sqrt{(x-a)(x-b)}=frac{x^2-(x-a)(x-b)}{x+sqrt{(x-a)(x-b)}}=
            frac{(a+b)-ab/x}{1+sqrt{(1-a/x)(1-b/x)}}$$
            for $x> max(a,b,0)$.






            share|cite|improve this answer









            $endgroup$



            You are correct, but the limit is easier to evaluate if you consider
            $$x-sqrt{(x-a)(x-b)}=frac{x^2-(x-a)(x-b)}{x+sqrt{(x-a)(x-b)}}=
            frac{(a+b)-ab/x}{1+sqrt{(1-a/x)(1-b/x)}}$$
            for $x> max(a,b,0)$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 5 at 13:41









            Robert ZRobert Z

            99.6k1068140




            99.6k1068140























                1












                $begingroup$

                Alternatively, do the change of variable $x = 1/t$
                $$lim_{xto+infty}x - sqrt{(x-a)(x-b)} = lim_{tto 0^+}frac{1 - sqrt{1 - (a + b)t + abt^2}}t$$
                and apply Taylor:
                $$sqrt{1 - (a + b)t + abt^2} = 1 - frac{a + b}2 t + cdots$$






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  Alternatively, do the change of variable $x = 1/t$
                  $$lim_{xto+infty}x - sqrt{(x-a)(x-b)} = lim_{tto 0^+}frac{1 - sqrt{1 - (a + b)t + abt^2}}t$$
                  and apply Taylor:
                  $$sqrt{1 - (a + b)t + abt^2} = 1 - frac{a + b}2 t + cdots$$






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    Alternatively, do the change of variable $x = 1/t$
                    $$lim_{xto+infty}x - sqrt{(x-a)(x-b)} = lim_{tto 0^+}frac{1 - sqrt{1 - (a + b)t + abt^2}}t$$
                    and apply Taylor:
                    $$sqrt{1 - (a + b)t + abt^2} = 1 - frac{a + b}2 t + cdots$$






                    share|cite|improve this answer









                    $endgroup$



                    Alternatively, do the change of variable $x = 1/t$
                    $$lim_{xto+infty}x - sqrt{(x-a)(x-b)} = lim_{tto 0^+}frac{1 - sqrt{1 - (a + b)t + abt^2}}t$$
                    and apply Taylor:
                    $$sqrt{1 - (a + b)t + abt^2} = 1 - frac{a + b}2 t + cdots$$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 5 at 15:13









                    Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla

                    34.5k42871




                    34.5k42871























                        1












                        $begingroup$

                        Hint:



                        Evaluation of the limit $displaystylelim_{xtoinfty}x-sqrt{(x-a)(x-b)}$,



                        Rationalizing the numerator,
                        $$lim_{tto0^+}dfrac{1-sqrt{1-(a+b)t+abt^2}}t=lim_{tto0^+}dfrac{1-(1-(a+b)t+abt^2)}tcdotlim_{tto0^+}dfrac1{1+sqrt{1-(a+b)t+abt^2}}$$



                        As $tto0,tne0,dfrac{1-(1-(a+b)t+abt^2)}t=(a+b)-abt$






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          Hint:



                          Evaluation of the limit $displaystylelim_{xtoinfty}x-sqrt{(x-a)(x-b)}$,



                          Rationalizing the numerator,
                          $$lim_{tto0^+}dfrac{1-sqrt{1-(a+b)t+abt^2}}t=lim_{tto0^+}dfrac{1-(1-(a+b)t+abt^2)}tcdotlim_{tto0^+}dfrac1{1+sqrt{1-(a+b)t+abt^2}}$$



                          As $tto0,tne0,dfrac{1-(1-(a+b)t+abt^2)}t=(a+b)-abt$






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            Hint:



                            Evaluation of the limit $displaystylelim_{xtoinfty}x-sqrt{(x-a)(x-b)}$,



                            Rationalizing the numerator,
                            $$lim_{tto0^+}dfrac{1-sqrt{1-(a+b)t+abt^2}}t=lim_{tto0^+}dfrac{1-(1-(a+b)t+abt^2)}tcdotlim_{tto0^+}dfrac1{1+sqrt{1-(a+b)t+abt^2}}$$



                            As $tto0,tne0,dfrac{1-(1-(a+b)t+abt^2)}t=(a+b)-abt$






                            share|cite|improve this answer









                            $endgroup$



                            Hint:



                            Evaluation of the limit $displaystylelim_{xtoinfty}x-sqrt{(x-a)(x-b)}$,



                            Rationalizing the numerator,
                            $$lim_{tto0^+}dfrac{1-sqrt{1-(a+b)t+abt^2}}t=lim_{tto0^+}dfrac{1-(1-(a+b)t+abt^2)}tcdotlim_{tto0^+}dfrac1{1+sqrt{1-(a+b)t+abt^2}}$$



                            As $tto0,tne0,dfrac{1-(1-(a+b)t+abt^2)}t=(a+b)-abt$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 5 at 16:52









                            lab bhattacharjeelab bhattacharjee

                            226k15157275




                            226k15157275






























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                                Mont Emei

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                                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...
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