What is the smallest value of the constant K












1












$begingroup$


Let $a, b$ are positive real numbers such that $a - b = 10$ , then the smallest value of the constant $K$ for which $sqrt{(x^2 + ax)} - sqrt {(x^2 +bx)} < K$ for all $x>0$.



My try :
enter image description here



Unable to solve further






real-analysis calculus limits optimization







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    Let $a, b$ are positive real numbers such that $a - b = 10$ , then the smallest value of the constant $K$ for which $sqrt{(x^2 + ax)} - sqrt {(x^2 +bx)} < K$ for all $x>0$.



    My try :
    enter image description here



    Unable to solve further






    real-analysis calculus limits optimization







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      1



      $begingroup$


      Let $a, b$ are positive real numbers such that $a - b = 10$ , then the smallest value of the constant $K$ for which $sqrt{(x^2 + ax)} - sqrt {(x^2 +bx)} < K$ for all $x>0$.



      My try :
      enter image description here



      Unable to solve further






      real-analysis calculus limits optimization







      share|cite|improve this question











      $endgroup$




      Let $a, b$ are positive real numbers such that $a - b = 10$ , then the smallest value of the constant $K$ for which $sqrt{(x^2 + ax)} - sqrt {(x^2 +bx)} < K$ for all $x>0$.



      My try :
      enter image description here



      Unable to solve further






      real-analysis calculus limits optimization




      real-analysis calculus limits optimization






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 16 at 18:24









      Maria Mazur

      49.7k1361124




      49.7k1361124










      asked Jan 16 at 15:51









      catttcattt

      242




      242






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          Hint.



          $$
          sqrt{(x^2 + ax)} - sqrt {(x^2 +bx)} = xleft(sqrt{frac{a}{x}+1}-sqrt{frac{b}{x}+1}right)
          $$



          and for large $x$



          $$
          sqrt{frac{a}{x}+1}-sqrt{frac{b}{x}+1} = frac{a-b}{2 x}+frac{1}{8} left(frac{1}{x}right)^2
          left(b^2-a^2right)+Oleft(left(frac{1}{x}right)^3right)
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Why do we have to consider large x ?
            $endgroup$
            – cattt
            Jan 17 at 2:25










          • $begingroup$
            @cattt Because for small $x$ the expression has a value near zero, being monotonically increasing as $x$ increases.
            $endgroup$
            – Cesareo
            Jan 17 at 8:48



















          1












          $begingroup$

          Using $$ sqrt{a}-sqrt{b} = {a-bover sqrt{a}+sqrt{b}}$$



          so $${(x^2+ax)-(x^2+bx)over sqrt{(x^2 + ax)} + sqrt {(x^2 +bx)}} < K$$



          thus $${10xover sqrt{(x^2 + ax)} + sqrt {(x^2 +bx)}} < K$$



          taking $xto infty$ we get $$lim _{xto infty}{10over sqrt{1 + {aover x}} + sqrt {1 +{bover x}}} = 5 leq K$$






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Shouldn't your last inequality be $leq$ instead of $<$ ?
            $endgroup$
            – Digitalis
            Jan 16 at 18:31












          Your Answer





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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          Hint.



          $$
          sqrt{(x^2 + ax)} - sqrt {(x^2 +bx)} = xleft(sqrt{frac{a}{x}+1}-sqrt{frac{b}{x}+1}right)
          $$



          and for large $x$



          $$
          sqrt{frac{a}{x}+1}-sqrt{frac{b}{x}+1} = frac{a-b}{2 x}+frac{1}{8} left(frac{1}{x}right)^2
          left(b^2-a^2right)+Oleft(left(frac{1}{x}right)^3right)
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Why do we have to consider large x ?
            $endgroup$
            – cattt
            Jan 17 at 2:25










          • $begingroup$
            @cattt Because for small $x$ the expression has a value near zero, being monotonically increasing as $x$ increases.
            $endgroup$
            – Cesareo
            Jan 17 at 8:48
















          1












          $begingroup$

          Hint.



          $$
          sqrt{(x^2 + ax)} - sqrt {(x^2 +bx)} = xleft(sqrt{frac{a}{x}+1}-sqrt{frac{b}{x}+1}right)
          $$



          and for large $x$



          $$
          sqrt{frac{a}{x}+1}-sqrt{frac{b}{x}+1} = frac{a-b}{2 x}+frac{1}{8} left(frac{1}{x}right)^2
          left(b^2-a^2right)+Oleft(left(frac{1}{x}right)^3right)
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Why do we have to consider large x ?
            $endgroup$
            – cattt
            Jan 17 at 2:25










          • $begingroup$
            @cattt Because for small $x$ the expression has a value near zero, being monotonically increasing as $x$ increases.
            $endgroup$
            – Cesareo
            Jan 17 at 8:48














          1












          1








          1





          $begingroup$

          Hint.



          $$
          sqrt{(x^2 + ax)} - sqrt {(x^2 +bx)} = xleft(sqrt{frac{a}{x}+1}-sqrt{frac{b}{x}+1}right)
          $$



          and for large $x$



          $$
          sqrt{frac{a}{x}+1}-sqrt{frac{b}{x}+1} = frac{a-b}{2 x}+frac{1}{8} left(frac{1}{x}right)^2
          left(b^2-a^2right)+Oleft(left(frac{1}{x}right)^3right)
          $$






          share|cite|improve this answer









          $endgroup$



          Hint.



          $$
          sqrt{(x^2 + ax)} - sqrt {(x^2 +bx)} = xleft(sqrt{frac{a}{x}+1}-sqrt{frac{b}{x}+1}right)
          $$



          and for large $x$



          $$
          sqrt{frac{a}{x}+1}-sqrt{frac{b}{x}+1} = frac{a-b}{2 x}+frac{1}{8} left(frac{1}{x}right)^2
          left(b^2-a^2right)+Oleft(left(frac{1}{x}right)^3right)
          $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 16 at 16:42









          CesareoCesareo

          9,7263517




          9,7263517












          • $begingroup$
            Why do we have to consider large x ?
            $endgroup$
            – cattt
            Jan 17 at 2:25










          • $begingroup$
            @cattt Because for small $x$ the expression has a value near zero, being monotonically increasing as $x$ increases.
            $endgroup$
            – Cesareo
            Jan 17 at 8:48


















          • $begingroup$
            Why do we have to consider large x ?
            $endgroup$
            – cattt
            Jan 17 at 2:25










          • $begingroup$
            @cattt Because for small $x$ the expression has a value near zero, being monotonically increasing as $x$ increases.
            $endgroup$
            – Cesareo
            Jan 17 at 8:48
















          $begingroup$
          Why do we have to consider large x ?
          $endgroup$
          – cattt
          Jan 17 at 2:25




          $begingroup$
          Why do we have to consider large x ?
          $endgroup$
          – cattt
          Jan 17 at 2:25












          $begingroup$
          @cattt Because for small $x$ the expression has a value near zero, being monotonically increasing as $x$ increases.
          $endgroup$
          – Cesareo
          Jan 17 at 8:48




          $begingroup$
          @cattt Because for small $x$ the expression has a value near zero, being monotonically increasing as $x$ increases.
          $endgroup$
          – Cesareo
          Jan 17 at 8:48











          1












          $begingroup$

          Using $$ sqrt{a}-sqrt{b} = {a-bover sqrt{a}+sqrt{b}}$$



          so $${(x^2+ax)-(x^2+bx)over sqrt{(x^2 + ax)} + sqrt {(x^2 +bx)}} < K$$



          thus $${10xover sqrt{(x^2 + ax)} + sqrt {(x^2 +bx)}} < K$$



          taking $xto infty$ we get $$lim _{xto infty}{10over sqrt{1 + {aover x}} + sqrt {1 +{bover x}}} = 5 leq K$$






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Shouldn't your last inequality be $leq$ instead of $<$ ?
            $endgroup$
            – Digitalis
            Jan 16 at 18:31
















          1












          $begingroup$

          Using $$ sqrt{a}-sqrt{b} = {a-bover sqrt{a}+sqrt{b}}$$



          so $${(x^2+ax)-(x^2+bx)over sqrt{(x^2 + ax)} + sqrt {(x^2 +bx)}} < K$$



          thus $${10xover sqrt{(x^2 + ax)} + sqrt {(x^2 +bx)}} < K$$



          taking $xto infty$ we get $$lim _{xto infty}{10over sqrt{1 + {aover x}} + sqrt {1 +{bover x}}} = 5 leq K$$






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Shouldn't your last inequality be $leq$ instead of $<$ ?
            $endgroup$
            – Digitalis
            Jan 16 at 18:31














          1












          1








          1





          $begingroup$

          Using $$ sqrt{a}-sqrt{b} = {a-bover sqrt{a}+sqrt{b}}$$



          so $${(x^2+ax)-(x^2+bx)over sqrt{(x^2 + ax)} + sqrt {(x^2 +bx)}} < K$$



          thus $${10xover sqrt{(x^2 + ax)} + sqrt {(x^2 +bx)}} < K$$



          taking $xto infty$ we get $$lim _{xto infty}{10over sqrt{1 + {aover x}} + sqrt {1 +{bover x}}} = 5 leq K$$






          share|cite|improve this answer











          $endgroup$



          Using $$ sqrt{a}-sqrt{b} = {a-bover sqrt{a}+sqrt{b}}$$



          so $${(x^2+ax)-(x^2+bx)over sqrt{(x^2 + ax)} + sqrt {(x^2 +bx)}} < K$$



          thus $${10xover sqrt{(x^2 + ax)} + sqrt {(x^2 +bx)}} < K$$



          taking $xto infty$ we get $$lim _{xto infty}{10over sqrt{1 + {aover x}} + sqrt {1 +{bover x}}} = 5 leq K$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 16 at 18:32

























          answered Jan 16 at 16:54









          Maria MazurMaria Mazur

          49.7k1361124




          49.7k1361124








          • 1




            $begingroup$
            Shouldn't your last inequality be $leq$ instead of $<$ ?
            $endgroup$
            – Digitalis
            Jan 16 at 18:31














          • 1




            $begingroup$
            Shouldn't your last inequality be $leq$ instead of $<$ ?
            $endgroup$
            – Digitalis
            Jan 16 at 18:31








          1




          1




          $begingroup$
          Shouldn't your last inequality be $leq$ instead of $<$ ?
          $endgroup$
          – Digitalis
          Jan 16 at 18:31




          $begingroup$
          Shouldn't your last inequality be $leq$ instead of $<$ ?
          $endgroup$
          – Digitalis
          Jan 16 at 18:31


















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