Evaluating $limlimits_{xtoinfty}int_0^1frac{ln x}{sqrt{x+t}}dt$












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Question: Evaluate the limit of this integral
enter image description here
Attempt:
enter image description here
I got the integral. but for some reason I can't find the limit at all. L'Hopital's doesn't even work






calculus integration limits







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    $begingroup$


    Question: Evaluate the limit of this integral
    enter image description here
    Attempt:
    enter image description here
    I got the integral. but for some reason I can't find the limit at all. L'Hopital's doesn't even work






    calculus integration limits







    share|cite|improve this question











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      2












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      2





      $begingroup$


      Question: Evaluate the limit of this integral
      enter image description here
      Attempt:
      enter image description here
      I got the integral. but for some reason I can't find the limit at all. L'Hopital's doesn't even work






      calculus integration limits







      share|cite|improve this question











      $endgroup$




      Question: Evaluate the limit of this integral
      enter image description here
      Attempt:
      enter image description here
      I got the integral. but for some reason I can't find the limit at all. L'Hopital's doesn't even work






      calculus integration limits




      calculus integration limits






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      edited yesterday









      Asaf Karagila♦

      305k33435766




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      asked yesterday









      I Main JayceI Main Jayce

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          2 Answers
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          $begingroup$

          Hint:$$lim_{x to infty} ln x int_{0}^{1}dfrac{1}{sqrt{t+x}}mathrm dt=2cdotlim_{x to infty}dfrac{sqrt{1+x}-sqrt{x}}{1/ln x}=2cdot lim_{xto infty}dfrac{ln x}{frac{1}{sqrt{1+x}-sqrt{x}}}$$




          $$dfrac{1}{sqrt{1+x}-sqrt{x}}cdotdfrac{sqrt{x+1}+sqrt{x}}{sqrt{x+1}+sqrt{x}}=sqrt{x+1}+sqrt{x} $$




          $$2cdotlim_{xto infty} dfrac{ln x}{sqrt{x+1}+sqrt{x}}=2lim_{xto infty}dfrac{1/x}{1/2cdotleft(frac{1}{sqrt{x+1}}+frac{1}{sqrt{x}}
          right)}$$






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          • $begingroup$
            Hi Paras, I've tried that. I rationalized the numerator and tried L'Hopital's. But it will just end on a loop. Could you please give further hints?
            $endgroup$
            – I Main Jayce
            yesterday



















          3












          $begingroup$

          $$ intdfrac{log( x)}{sqrt{t+x}}, dt=2 sqrt{t+x} log (x)$$
          $$ int_{0}^{1}dfrac{log( x)}{sqrt{t+x}}, dt=2 left(sqrt{x+1}-sqrt{x}right) log (x)=2sqrt{x}left(sqrt{1+frac{1}{x}}-1 right)log(x)$$



          Now, since $x$ is large
          $$sqrt{1+frac{1}{x}}=1+frac{1}{2 x}+Oleft(frac{1}{x^2}right)$$
          $$2sqrt{x}left(sqrt{1+frac{1}{x}}-1 right)log(x)=2sqrt{x}left(frac{1}{2 x}+Oleft(frac{1}{x^2}right) right)log(x)simfrac{log(x)}{sqrt x}= 2frac{log(sqrt x)}{sqrt x}=2frac {log(y)}y$$



          Try with $x=100$; the exact result would be $2 left(sqrt{101}-10right) log (100)approx 0.459371$ while the approximation gives $frac{log (10)}{5}approx 0.460517$.






          share|cite|improve this answer









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            2 Answers
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            2 Answers
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            $begingroup$

            Hint:$$lim_{x to infty} ln x int_{0}^{1}dfrac{1}{sqrt{t+x}}mathrm dt=2cdotlim_{x to infty}dfrac{sqrt{1+x}-sqrt{x}}{1/ln x}=2cdot lim_{xto infty}dfrac{ln x}{frac{1}{sqrt{1+x}-sqrt{x}}}$$




            $$dfrac{1}{sqrt{1+x}-sqrt{x}}cdotdfrac{sqrt{x+1}+sqrt{x}}{sqrt{x+1}+sqrt{x}}=sqrt{x+1}+sqrt{x} $$




            $$2cdotlim_{xto infty} dfrac{ln x}{sqrt{x+1}+sqrt{x}}=2lim_{xto infty}dfrac{1/x}{1/2cdotleft(frac{1}{sqrt{x+1}}+frac{1}{sqrt{x}}
            right)}$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Hi Paras, I've tried that. I rationalized the numerator and tried L'Hopital's. But it will just end on a loop. Could you please give further hints?
              $endgroup$
              – I Main Jayce
              yesterday
















            3












            $begingroup$

            Hint:$$lim_{x to infty} ln x int_{0}^{1}dfrac{1}{sqrt{t+x}}mathrm dt=2cdotlim_{x to infty}dfrac{sqrt{1+x}-sqrt{x}}{1/ln x}=2cdot lim_{xto infty}dfrac{ln x}{frac{1}{sqrt{1+x}-sqrt{x}}}$$




            $$dfrac{1}{sqrt{1+x}-sqrt{x}}cdotdfrac{sqrt{x+1}+sqrt{x}}{sqrt{x+1}+sqrt{x}}=sqrt{x+1}+sqrt{x} $$




            $$2cdotlim_{xto infty} dfrac{ln x}{sqrt{x+1}+sqrt{x}}=2lim_{xto infty}dfrac{1/x}{1/2cdotleft(frac{1}{sqrt{x+1}}+frac{1}{sqrt{x}}
            right)}$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Hi Paras, I've tried that. I rationalized the numerator and tried L'Hopital's. But it will just end on a loop. Could you please give further hints?
              $endgroup$
              – I Main Jayce
              yesterday














            3












            3








            3





            $begingroup$

            Hint:$$lim_{x to infty} ln x int_{0}^{1}dfrac{1}{sqrt{t+x}}mathrm dt=2cdotlim_{x to infty}dfrac{sqrt{1+x}-sqrt{x}}{1/ln x}=2cdot lim_{xto infty}dfrac{ln x}{frac{1}{sqrt{1+x}-sqrt{x}}}$$




            $$dfrac{1}{sqrt{1+x}-sqrt{x}}cdotdfrac{sqrt{x+1}+sqrt{x}}{sqrt{x+1}+sqrt{x}}=sqrt{x+1}+sqrt{x} $$




            $$2cdotlim_{xto infty} dfrac{ln x}{sqrt{x+1}+sqrt{x}}=2lim_{xto infty}dfrac{1/x}{1/2cdotleft(frac{1}{sqrt{x+1}}+frac{1}{sqrt{x}}
            right)}$$






            share|cite|improve this answer











            $endgroup$



            Hint:$$lim_{x to infty} ln x int_{0}^{1}dfrac{1}{sqrt{t+x}}mathrm dt=2cdotlim_{x to infty}dfrac{sqrt{1+x}-sqrt{x}}{1/ln x}=2cdot lim_{xto infty}dfrac{ln x}{frac{1}{sqrt{1+x}-sqrt{x}}}$$




            $$dfrac{1}{sqrt{1+x}-sqrt{x}}cdotdfrac{sqrt{x+1}+sqrt{x}}{sqrt{x+1}+sqrt{x}}=sqrt{x+1}+sqrt{x} $$




            $$2cdotlim_{xto infty} dfrac{ln x}{sqrt{x+1}+sqrt{x}}=2lim_{xto infty}dfrac{1/x}{1/2cdotleft(frac{1}{sqrt{x+1}}+frac{1}{sqrt{x}}
            right)}$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited yesterday

























            answered yesterday









            Paras KhoslaParas Khosla

            1,385219




            1,385219












            • $begingroup$
              Hi Paras, I've tried that. I rationalized the numerator and tried L'Hopital's. But it will just end on a loop. Could you please give further hints?
              $endgroup$
              – I Main Jayce
              yesterday


















            • $begingroup$
              Hi Paras, I've tried that. I rationalized the numerator and tried L'Hopital's. But it will just end on a loop. Could you please give further hints?
              $endgroup$
              – I Main Jayce
              yesterday
















            $begingroup$
            Hi Paras, I've tried that. I rationalized the numerator and tried L'Hopital's. But it will just end on a loop. Could you please give further hints?
            $endgroup$
            – I Main Jayce
            yesterday




            $begingroup$
            Hi Paras, I've tried that. I rationalized the numerator and tried L'Hopital's. But it will just end on a loop. Could you please give further hints?
            $endgroup$
            – I Main Jayce
            yesterday











            3












            $begingroup$

            $$ intdfrac{log( x)}{sqrt{t+x}}, dt=2 sqrt{t+x} log (x)$$
            $$ int_{0}^{1}dfrac{log( x)}{sqrt{t+x}}, dt=2 left(sqrt{x+1}-sqrt{x}right) log (x)=2sqrt{x}left(sqrt{1+frac{1}{x}}-1 right)log(x)$$



            Now, since $x$ is large
            $$sqrt{1+frac{1}{x}}=1+frac{1}{2 x}+Oleft(frac{1}{x^2}right)$$
            $$2sqrt{x}left(sqrt{1+frac{1}{x}}-1 right)log(x)=2sqrt{x}left(frac{1}{2 x}+Oleft(frac{1}{x^2}right) right)log(x)simfrac{log(x)}{sqrt x}= 2frac{log(sqrt x)}{sqrt x}=2frac {log(y)}y$$



            Try with $x=100$; the exact result would be $2 left(sqrt{101}-10right) log (100)approx 0.459371$ while the approximation gives $frac{log (10)}{5}approx 0.460517$.






            share|cite|improve this answer









            $endgroup$


















              3












              $begingroup$

              $$ intdfrac{log( x)}{sqrt{t+x}}, dt=2 sqrt{t+x} log (x)$$
              $$ int_{0}^{1}dfrac{log( x)}{sqrt{t+x}}, dt=2 left(sqrt{x+1}-sqrt{x}right) log (x)=2sqrt{x}left(sqrt{1+frac{1}{x}}-1 right)log(x)$$



              Now, since $x$ is large
              $$sqrt{1+frac{1}{x}}=1+frac{1}{2 x}+Oleft(frac{1}{x^2}right)$$
              $$2sqrt{x}left(sqrt{1+frac{1}{x}}-1 right)log(x)=2sqrt{x}left(frac{1}{2 x}+Oleft(frac{1}{x^2}right) right)log(x)simfrac{log(x)}{sqrt x}= 2frac{log(sqrt x)}{sqrt x}=2frac {log(y)}y$$



              Try with $x=100$; the exact result would be $2 left(sqrt{101}-10right) log (100)approx 0.459371$ while the approximation gives $frac{log (10)}{5}approx 0.460517$.






              share|cite|improve this answer









              $endgroup$
















                3












                3








                3





                $begingroup$

                $$ intdfrac{log( x)}{sqrt{t+x}}, dt=2 sqrt{t+x} log (x)$$
                $$ int_{0}^{1}dfrac{log( x)}{sqrt{t+x}}, dt=2 left(sqrt{x+1}-sqrt{x}right) log (x)=2sqrt{x}left(sqrt{1+frac{1}{x}}-1 right)log(x)$$



                Now, since $x$ is large
                $$sqrt{1+frac{1}{x}}=1+frac{1}{2 x}+Oleft(frac{1}{x^2}right)$$
                $$2sqrt{x}left(sqrt{1+frac{1}{x}}-1 right)log(x)=2sqrt{x}left(frac{1}{2 x}+Oleft(frac{1}{x^2}right) right)log(x)simfrac{log(x)}{sqrt x}= 2frac{log(sqrt x)}{sqrt x}=2frac {log(y)}y$$



                Try with $x=100$; the exact result would be $2 left(sqrt{101}-10right) log (100)approx 0.459371$ while the approximation gives $frac{log (10)}{5}approx 0.460517$.






                share|cite|improve this answer









                $endgroup$



                $$ intdfrac{log( x)}{sqrt{t+x}}, dt=2 sqrt{t+x} log (x)$$
                $$ int_{0}^{1}dfrac{log( x)}{sqrt{t+x}}, dt=2 left(sqrt{x+1}-sqrt{x}right) log (x)=2sqrt{x}left(sqrt{1+frac{1}{x}}-1 right)log(x)$$



                Now, since $x$ is large
                $$sqrt{1+frac{1}{x}}=1+frac{1}{2 x}+Oleft(frac{1}{x^2}right)$$
                $$2sqrt{x}left(sqrt{1+frac{1}{x}}-1 right)log(x)=2sqrt{x}left(frac{1}{2 x}+Oleft(frac{1}{x^2}right) right)log(x)simfrac{log(x)}{sqrt x}= 2frac{log(sqrt x)}{sqrt x}=2frac {log(y)}y$$



                Try with $x=100$; the exact result would be $2 left(sqrt{101}-10right) log (100)approx 0.459371$ while the approximation gives $frac{log (10)}{5}approx 0.460517$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered yesterday









                Claude LeiboviciClaude Leibovici

                123k1157134




                123k1157134






























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