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A problem on epsilon and delta in the proof of L'Hospital's Rule


What is the Least Upper bound on $delta$ in the $epsilon-delta$ definition of limit?Darboux Integrability epsilon-delta proofProof on showing F(x,y) is continuous by $epsilon - delta$ definitionepsilon delta approach to a problemDoubt in epsilon-delta limit proof (more specifically, the inequalities)Question about proof using epsilon delta for limitsDependence and Independence of $epsilon$ and $delta$Proving the limit of trigonometric function using epsilon delta definition.Epsilon-delta definitions, inequality strict / non-strict?Epsilon-delta derivative proof of $x^n$













1












$begingroup$


I'm following the proof of L'Hospital's Rule from a text but have trouble understanding this part regarding epsilon-delta.




Assume $Lgt 0$ and $epsilon gt 0$ is given and $delta in (0,1)$. What we have is



$$(L-epsilon)(1-delta)-deltalt fracf(alpha)g(alpha)lt(L+epsilon)+delta$$



If we take $delta=operatornamemin1,epsilon,epsilon/($ then
$$L-2epsilonle fracf(alpha)g(alpha)le L+2epsilon$$




I can't get the first inequality above. Also, shouldn't we get a strict inequality on the right side? I've tried some manipulations but all have been unsuccessful in getting that. How can I show this part?










share|cite|improve this question











$endgroup$
















    1












    $begingroup$


    I'm following the proof of L'Hospital's Rule from a text but have trouble understanding this part regarding epsilon-delta.




    Assume $Lgt 0$ and $epsilon gt 0$ is given and $delta in (0,1)$. What we have is



    $$(L-epsilon)(1-delta)-deltalt fracf(alpha)g(alpha)lt(L+epsilon)+delta$$



    If we take $delta=operatornamemin1,epsilon,epsilon/($ then
    $$L-2epsilonle fracf(alpha)g(alpha)le L+2epsilon$$




    I can't get the first inequality above. Also, shouldn't we get a strict inequality on the right side? I've tried some manipulations but all have been unsuccessful in getting that. How can I show this part?










    share|cite|improve this question











    $endgroup$














      1












      1








      1


      1



      $begingroup$


      I'm following the proof of L'Hospital's Rule from a text but have trouble understanding this part regarding epsilon-delta.




      Assume $Lgt 0$ and $epsilon gt 0$ is given and $delta in (0,1)$. What we have is



      $$(L-epsilon)(1-delta)-deltalt fracf(alpha)g(alpha)lt(L+epsilon)+delta$$



      If we take $delta=operatornamemin1,epsilon,epsilon/($ then
      $$L-2epsilonle fracf(alpha)g(alpha)le L+2epsilon$$




      I can't get the first inequality above. Also, shouldn't we get a strict inequality on the right side? I've tried some manipulations but all have been unsuccessful in getting that. How can I show this part?










      share|cite|improve this question











      $endgroup$




      I'm following the proof of L'Hospital's Rule from a text but have trouble understanding this part regarding epsilon-delta.




      Assume $Lgt 0$ and $epsilon gt 0$ is given and $delta in (0,1)$. What we have is



      $$(L-epsilon)(1-delta)-deltalt fracf(alpha)g(alpha)lt(L+epsilon)+delta$$



      If we take $delta=operatornamemin1,epsilon,epsilon/($ then
      $$L-2epsilonle fracf(alpha)g(alpha)le L+2epsilon$$




      I can't get the first inequality above. Also, shouldn't we get a strict inequality on the right side? I've tried some manipulations but all have been unsuccessful in getting that. How can I show this part?







      real-analysis analysis






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 14 at 7:13









      Winfield Chen

      484




      484










      asked Mar 13 '15 at 0:23









      takecaretakecare

      2,34021537




      2,34021537




















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