Is $displaystylelim _{xto infty }left(dfrac{x^2+2x+1}{x^2-3x+2}right)^x = e^5$? The 2019 Stack...

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Is $displaystylelim _{xto infty }left(dfrac{x^2+2x+1}{x^2-3x+2}right)^x = e^5$?



The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Showing $lim _{nrightarrow infty } frac {S_{n}}{n^{2}} = frac{1}{6}$Is there a form $lim_{xtoinfty}left[frac{infty}{infty}right] $ of l'Hôpital's rule?Evaluate $limlimits_{x to 0+} fleft(frac1xright) = lim limits_{x to infty} f(x)$Find $displaystylesum_{k=1}^inftydfrac{1}{left(binom{2014+k}{2014}right)}$Calculate $limlimits_{xtoinfty}left(frac{x-2}{x+2}right)^{3x}$evaluate lim: $lim _{xto infty }left(2xleft(e-left(1+frac{1}{x}right)^xright)right)$Evaluating $limlimits_{xtoinfty}xleft(dfrac{1}{e}-left(dfrac{x}{x+1}right)^xright)$Evaluate:$limlimits_{ntoinfty}sum_{k=0}^n frac1{k!}left(-frac12right)^k$How does one evaluate $lim _{nto infty }left(sqrt[n]{int _0^1:left(1+x^nright)^ndx}right)$?Calculate the limit: $lim_{ntoinfty}n^2left(left(1+frac{1}{n}right)^8-left(1+frac{2}{n}right)^4right)$












-1












$begingroup$


I am trying to evaluate
$displaystylelim _{xto infty }left(dfrac{x^2+2x+1}{x^2-3x+2}right)^x$.
I'm getting the answer $e^5$ using an online caculator. Can anyone suggest a formal solution?






limits







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    It will be nice if you show your solution
    $endgroup$
    – haqnatural
    Oct 22 '17 at 16:09










  • $begingroup$
    Thanks @haqnatural
    $endgroup$
    – Paras Khosla
    Oct 22 '17 at 16:10










  • $begingroup$
    this is the right answer, how do you proved it?
    $endgroup$
    – Dr. Sonnhard Graubner
    Oct 22 '17 at 16:10










  • $begingroup$
    Got the answer from an online calc
    $endgroup$
    – Paras Khosla
    Oct 22 '17 at 16:10










  • $begingroup$
    expressed the function using e as the base
    $endgroup$
    – Paras Khosla
    Oct 22 '17 at 16:10
















-1












$begingroup$


I am trying to evaluate
$displaystylelim _{xto infty }left(dfrac{x^2+2x+1}{x^2-3x+2}right)^x$.
I'm getting the answer $e^5$ using an online caculator. Can anyone suggest a formal solution?






limits







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    It will be nice if you show your solution
    $endgroup$
    – haqnatural
    Oct 22 '17 at 16:09










  • $begingroup$
    Thanks @haqnatural
    $endgroup$
    – Paras Khosla
    Oct 22 '17 at 16:10










  • $begingroup$
    this is the right answer, how do you proved it?
    $endgroup$
    – Dr. Sonnhard Graubner
    Oct 22 '17 at 16:10










  • $begingroup$
    Got the answer from an online calc
    $endgroup$
    – Paras Khosla
    Oct 22 '17 at 16:10










  • $begingroup$
    expressed the function using e as the base
    $endgroup$
    – Paras Khosla
    Oct 22 '17 at 16:10














-1












-1








-1





$begingroup$


I am trying to evaluate
$displaystylelim _{xto infty }left(dfrac{x^2+2x+1}{x^2-3x+2}right)^x$.
I'm getting the answer $e^5$ using an online caculator. Can anyone suggest a formal solution?






limits







share|cite|improve this question











$endgroup$




I am trying to evaluate
$displaystylelim _{xto infty }left(dfrac{x^2+2x+1}{x^2-3x+2}right)^x$.
I'm getting the answer $e^5$ using an online caculator. Can anyone suggest a formal solution?






limits




limits






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Oct 22 '17 at 16:19









venrey

16011




16011










asked Oct 22 '17 at 16:08









Paras KhoslaParas Khosla

3,257627




3,257627








  • 1




    $begingroup$
    It will be nice if you show your solution
    $endgroup$
    – haqnatural
    Oct 22 '17 at 16:09










  • $begingroup$
    Thanks @haqnatural
    $endgroup$
    – Paras Khosla
    Oct 22 '17 at 16:10










  • $begingroup$
    this is the right answer, how do you proved it?
    $endgroup$
    – Dr. Sonnhard Graubner
    Oct 22 '17 at 16:10










  • $begingroup$
    Got the answer from an online calc
    $endgroup$
    – Paras Khosla
    Oct 22 '17 at 16:10










  • $begingroup$
    expressed the function using e as the base
    $endgroup$
    – Paras Khosla
    Oct 22 '17 at 16:10














  • 1




    $begingroup$
    It will be nice if you show your solution
    $endgroup$
    – haqnatural
    Oct 22 '17 at 16:09










  • $begingroup$
    Thanks @haqnatural
    $endgroup$
    – Paras Khosla
    Oct 22 '17 at 16:10










  • $begingroup$
    this is the right answer, how do you proved it?
    $endgroup$
    – Dr. Sonnhard Graubner
    Oct 22 '17 at 16:10










  • $begingroup$
    Got the answer from an online calc
    $endgroup$
    – Paras Khosla
    Oct 22 '17 at 16:10










  • $begingroup$
    expressed the function using e as the base
    $endgroup$
    – Paras Khosla
    Oct 22 '17 at 16:10








1




1




$begingroup$
It will be nice if you show your solution
$endgroup$
– haqnatural
Oct 22 '17 at 16:09




$begingroup$
It will be nice if you show your solution
$endgroup$
– haqnatural
Oct 22 '17 at 16:09












$begingroup$
Thanks @haqnatural
$endgroup$
– Paras Khosla
Oct 22 '17 at 16:10




$begingroup$
Thanks @haqnatural
$endgroup$
– Paras Khosla
Oct 22 '17 at 16:10












$begingroup$
this is the right answer, how do you proved it?
$endgroup$
– Dr. Sonnhard Graubner
Oct 22 '17 at 16:10




$begingroup$
this is the right answer, how do you proved it?
$endgroup$
– Dr. Sonnhard Graubner
Oct 22 '17 at 16:10












$begingroup$
Got the answer from an online calc
$endgroup$
– Paras Khosla
Oct 22 '17 at 16:10




$begingroup$
Got the answer from an online calc
$endgroup$
– Paras Khosla
Oct 22 '17 at 16:10












$begingroup$
expressed the function using e as the base
$endgroup$
– Paras Khosla
Oct 22 '17 at 16:10




$begingroup$
expressed the function using e as the base
$endgroup$
– Paras Khosla
Oct 22 '17 at 16:10










2 Answers
2






active

oldest

votes


















3












$begingroup$

Hint:



$$left(left(frac{left(x^2+2x+1right)}{left(x^2-3x+2right)}right)^xright)$$



$$=dfrac{((1+x)^x)^2}{(x-1)^x(x-2)^x}$$



$$=dfrac{left(left(1+dfrac1xright)^xright)^2}{left(1-dfrac1xright)^xleft(1-dfrac2xright)^x}$$



Now use $$lim_{ntoinfty}left(1+dfrac1nright)^n=e$$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Well i know it has been more than a year but, as the problem is still open , i am writing the general methodology.



    Observe
    $displaystylelim _{xto infty }left(dfrac{x^2+2x+1}{x^2-3x+2}right)=1$
    And thus it is in $(1)^{infty}$ form
    Hence for the limit of this form answer is given by
    $$e^{lim _{xto infty }(f(x)-1)(g(x))}$$ where $lim _{xto infty } f(x)^{g(x)}$ must tend to the $(1)^{infty}$ form.



    Rest is just calculation.






    share|cite|improve this answer











    $endgroup$














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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      Hint:



      $$left(left(frac{left(x^2+2x+1right)}{left(x^2-3x+2right)}right)^xright)$$



      $$=dfrac{((1+x)^x)^2}{(x-1)^x(x-2)^x}$$



      $$=dfrac{left(left(1+dfrac1xright)^xright)^2}{left(1-dfrac1xright)^xleft(1-dfrac2xright)^x}$$



      Now use $$lim_{ntoinfty}left(1+dfrac1nright)^n=e$$






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        Hint:



        $$left(left(frac{left(x^2+2x+1right)}{left(x^2-3x+2right)}right)^xright)$$



        $$=dfrac{((1+x)^x)^2}{(x-1)^x(x-2)^x}$$



        $$=dfrac{left(left(1+dfrac1xright)^xright)^2}{left(1-dfrac1xright)^xleft(1-dfrac2xright)^x}$$



        Now use $$lim_{ntoinfty}left(1+dfrac1nright)^n=e$$






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          Hint:



          $$left(left(frac{left(x^2+2x+1right)}{left(x^2-3x+2right)}right)^xright)$$



          $$=dfrac{((1+x)^x)^2}{(x-1)^x(x-2)^x}$$



          $$=dfrac{left(left(1+dfrac1xright)^xright)^2}{left(1-dfrac1xright)^xleft(1-dfrac2xright)^x}$$



          Now use $$lim_{ntoinfty}left(1+dfrac1nright)^n=e$$






          share|cite|improve this answer









          $endgroup$



          Hint:



          $$left(left(frac{left(x^2+2x+1right)}{left(x^2-3x+2right)}right)^xright)$$



          $$=dfrac{((1+x)^x)^2}{(x-1)^x(x-2)^x}$$



          $$=dfrac{left(left(1+dfrac1xright)^xright)^2}{left(1-dfrac1xright)^xleft(1-dfrac2xright)^x}$$



          Now use $$lim_{ntoinfty}left(1+dfrac1nright)^n=e$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Oct 22 '17 at 16:12









          lab bhattacharjeelab bhattacharjee

          228k15159279




          228k15159279























              1












              $begingroup$

              Well i know it has been more than a year but, as the problem is still open , i am writing the general methodology.



              Observe
              $displaystylelim _{xto infty }left(dfrac{x^2+2x+1}{x^2-3x+2}right)=1$
              And thus it is in $(1)^{infty}$ form
              Hence for the limit of this form answer is given by
              $$e^{lim _{xto infty }(f(x)-1)(g(x))}$$ where $lim _{xto infty } f(x)^{g(x)}$ must tend to the $(1)^{infty}$ form.



              Rest is just calculation.






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                Well i know it has been more than a year but, as the problem is still open , i am writing the general methodology.



                Observe
                $displaystylelim _{xto infty }left(dfrac{x^2+2x+1}{x^2-3x+2}right)=1$
                And thus it is in $(1)^{infty}$ form
                Hence for the limit of this form answer is given by
                $$e^{lim _{xto infty }(f(x)-1)(g(x))}$$ where $lim _{xto infty } f(x)^{g(x)}$ must tend to the $(1)^{infty}$ form.



                Rest is just calculation.






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Well i know it has been more than a year but, as the problem is still open , i am writing the general methodology.



                  Observe
                  $displaystylelim _{xto infty }left(dfrac{x^2+2x+1}{x^2-3x+2}right)=1$
                  And thus it is in $(1)^{infty}$ form
                  Hence for the limit of this form answer is given by
                  $$e^{lim _{xto infty }(f(x)-1)(g(x))}$$ where $lim _{xto infty } f(x)^{g(x)}$ must tend to the $(1)^{infty}$ form.



                  Rest is just calculation.






                  share|cite|improve this answer











                  $endgroup$



                  Well i know it has been more than a year but, as the problem is still open , i am writing the general methodology.



                  Observe
                  $displaystylelim _{xto infty }left(dfrac{x^2+2x+1}{x^2-3x+2}right)=1$
                  And thus it is in $(1)^{infty}$ form
                  Hence for the limit of this form answer is given by
                  $$e^{lim _{xto infty }(f(x)-1)(g(x))}$$ where $lim _{xto infty } f(x)^{g(x)}$ must tend to the $(1)^{infty}$ form.



                  Rest is just calculation.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Mar 22 at 20:37

























                  answered Mar 22 at 20:30









                  NewBornMATHNewBornMATH

                  549111




                  549111






























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