Why isn't L Hospital Rule applicable here?












2














$lim_{x to 0} frac{e^frac{1}{x}}{e^frac{1}{x}+1}$



I'm trying to use this form of LH Rule. enter image description here






calculus







share|cite|improve this question






















  • Why not? I think it is allright
    – José Alejandro Aburto Araneda
    5 hours ago






  • 6




    The problem is that $e^{1/x}$ doesn't approach a limit as $xto 0$. It approaches $+infty$ as $x$ goes to $0$ from the right, and it goes to $0$ as $x$ goes to $0$ from the left. Thus your expression goes to $1$ on the right, and $0$ on the left.
    – lulu
    5 hours ago












  • You can use L'hopital to find the right side limit but it's not necessary. You can't apply it to find the left side limit as that is neither $infty/infty$ nor $0/0$.
    – fleablood
    4 hours ago










  • @fleablood Oh okay! The theorem is only about the right hand limit. Thanks!
    – user1752323
    1 hour ago
















2














$lim_{x to 0} frac{e^frac{1}{x}}{e^frac{1}{x}+1}$



I'm trying to use this form of LH Rule. enter image description here






calculus







share|cite|improve this question






















  • Why not? I think it is allright
    – José Alejandro Aburto Araneda
    5 hours ago






  • 6




    The problem is that $e^{1/x}$ doesn't approach a limit as $xto 0$. It approaches $+infty$ as $x$ goes to $0$ from the right, and it goes to $0$ as $x$ goes to $0$ from the left. Thus your expression goes to $1$ on the right, and $0$ on the left.
    – lulu
    5 hours ago












  • You can use L'hopital to find the right side limit but it's not necessary. You can't apply it to find the left side limit as that is neither $infty/infty$ nor $0/0$.
    – fleablood
    4 hours ago










  • @fleablood Oh okay! The theorem is only about the right hand limit. Thanks!
    – user1752323
    1 hour ago














2












2








2







$lim_{x to 0} frac{e^frac{1}{x}}{e^frac{1}{x}+1}$



I'm trying to use this form of LH Rule. enter image description here






calculus







share|cite|improve this question













$lim_{x to 0} frac{e^frac{1}{x}}{e^frac{1}{x}+1}$



I'm trying to use this form of LH Rule. enter image description here






calculus




calculus






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 5 hours ago









user1752323

324




324












  • Why not? I think it is allright
    – José Alejandro Aburto Araneda
    5 hours ago






  • 6




    The problem is that $e^{1/x}$ doesn't approach a limit as $xto 0$. It approaches $+infty$ as $x$ goes to $0$ from the right, and it goes to $0$ as $x$ goes to $0$ from the left. Thus your expression goes to $1$ on the right, and $0$ on the left.
    – lulu
    5 hours ago












  • You can use L'hopital to find the right side limit but it's not necessary. You can't apply it to find the left side limit as that is neither $infty/infty$ nor $0/0$.
    – fleablood
    4 hours ago










  • @fleablood Oh okay! The theorem is only about the right hand limit. Thanks!
    – user1752323
    1 hour ago


















  • Why not? I think it is allright
    – José Alejandro Aburto Araneda
    5 hours ago






  • 6




    The problem is that $e^{1/x}$ doesn't approach a limit as $xto 0$. It approaches $+infty$ as $x$ goes to $0$ from the right, and it goes to $0$ as $x$ goes to $0$ from the left. Thus your expression goes to $1$ on the right, and $0$ on the left.
    – lulu
    5 hours ago












  • You can use L'hopital to find the right side limit but it's not necessary. You can't apply it to find the left side limit as that is neither $infty/infty$ nor $0/0$.
    – fleablood
    4 hours ago










  • @fleablood Oh okay! The theorem is only about the right hand limit. Thanks!
    – user1752323
    1 hour ago
















Why not? I think it is allright
– José Alejandro Aburto Araneda
5 hours ago




Why not? I think it is allright
– José Alejandro Aburto Araneda
5 hours ago




6




6




The problem is that $e^{1/x}$ doesn't approach a limit as $xto 0$. It approaches $+infty$ as $x$ goes to $0$ from the right, and it goes to $0$ as $x$ goes to $0$ from the left. Thus your expression goes to $1$ on the right, and $0$ on the left.
– lulu
5 hours ago






The problem is that $e^{1/x}$ doesn't approach a limit as $xto 0$. It approaches $+infty$ as $x$ goes to $0$ from the right, and it goes to $0$ as $x$ goes to $0$ from the left. Thus your expression goes to $1$ on the right, and $0$ on the left.
– lulu
5 hours ago














You can use L'hopital to find the right side limit but it's not necessary. You can't apply it to find the left side limit as that is neither $infty/infty$ nor $0/0$.
– fleablood
4 hours ago




You can use L'hopital to find the right side limit but it's not necessary. You can't apply it to find the left side limit as that is neither $infty/infty$ nor $0/0$.
– fleablood
4 hours ago












@fleablood Oh okay! The theorem is only about the right hand limit. Thanks!
– user1752323
1 hour ago




@fleablood Oh okay! The theorem is only about the right hand limit. Thanks!
– user1752323
1 hour ago










3 Answers
3






active

oldest

votes


















3














You can apply l’Hôpital for the limit from the right, because it is in the form $infty/infty$.



The limit from the left is instead in the form $0/1$, which is not indeterminate. In order to use that form of l'Hôpital for a two sided limit, you need $infty/infty$ on both sides.



More generally you need $infty/infty$ or $0/0$ on either side.



If you blindly apply l’Hôpital, you end up with
$$
lim_{xto0}frac{-e^{1/x}/x^2}{-e^{1/x}/x^2}=1
$$
but this would be incorrect, because
$$
lim_{xto0^-}frac{e^{1/x}}{e^{1/x}+1}=0
$$
For the limit from the right the limit is correct, but using l’Hôpital is of course not necessary:
$$
lim_{xto0^+}frac{e^{1/x}}{e^{1/x}+1}=lim_{xto0^+}frac{1}{1+e^{-1/x}}=
frac{1}{1+0}=1
$$






share|cite|improve this answer































    3














    L'Hopital's rule does not apply because the limit
    $$
    lim_{xto 0}e^{1/x}
    $$
    does not exist, since
    $$
    lim_{xto 0^-}e^{1/x}=0quadtext{while}quad lim_{xto 0^+}e^{1/x}=infty
    $$
    So, consequently
    $$
    lim_{xto 0^-}frac{e^{1/x}}{e^{1/x}+1}=0quadtext{while}quad lim_{xto 0^+}frac{e^{1/x}}{e^{1/x}+1}=1
    $$






    share|cite|improve this answer



















    • 2




      This is not the reason why l'Hôpital cannot be applied.
      – egreg
      4 hours ago



















    0














    here, $$when~~~xto 0^+,e^{frac{1}{x}}to infty$$
    So,$$lim_{x to 0^+} frac{e^frac{1}{x}}{e^frac{1}{x}+1}=lim_{x to 0^+}frac{e^frac{1}{x}}{e^frac{1}{x}left(1+frac{1}{e^frac{1}{x}} right)}=lim_{x to 0^+}frac{1}{left(1+frac{1}{e^frac{1}{x}} right)}=frac{1}{1+0}=1$$
    $$when~~~xto 0^-,e^{frac{1}{x}} to 0$$
    So,
    $$lim_{x to 0^-} frac{e^frac{1}{x}}{e^frac{1}{x}+1}=frac{0}{0+1}=0$$



    so,simply limit doesn't exist.so we can't find the limit when $x to 0$ no matter what procedure you follow.






    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

      votes









      3














      You can apply l’Hôpital for the limit from the right, because it is in the form $infty/infty$.



      The limit from the left is instead in the form $0/1$, which is not indeterminate. In order to use that form of l'Hôpital for a two sided limit, you need $infty/infty$ on both sides.



      More generally you need $infty/infty$ or $0/0$ on either side.



      If you blindly apply l’Hôpital, you end up with
      $$
      lim_{xto0}frac{-e^{1/x}/x^2}{-e^{1/x}/x^2}=1
      $$
      but this would be incorrect, because
      $$
      lim_{xto0^-}frac{e^{1/x}}{e^{1/x}+1}=0
      $$
      For the limit from the right the limit is correct, but using l’Hôpital is of course not necessary:
      $$
      lim_{xto0^+}frac{e^{1/x}}{e^{1/x}+1}=lim_{xto0^+}frac{1}{1+e^{-1/x}}=
      frac{1}{1+0}=1
      $$






      share|cite|improve this answer




























        3














        You can apply l’Hôpital for the limit from the right, because it is in the form $infty/infty$.



        The limit from the left is instead in the form $0/1$, which is not indeterminate. In order to use that form of l'Hôpital for a two sided limit, you need $infty/infty$ on both sides.



        More generally you need $infty/infty$ or $0/0$ on either side.



        If you blindly apply l’Hôpital, you end up with
        $$
        lim_{xto0}frac{-e^{1/x}/x^2}{-e^{1/x}/x^2}=1
        $$
        but this would be incorrect, because
        $$
        lim_{xto0^-}frac{e^{1/x}}{e^{1/x}+1}=0
        $$
        For the limit from the right the limit is correct, but using l’Hôpital is of course not necessary:
        $$
        lim_{xto0^+}frac{e^{1/x}}{e^{1/x}+1}=lim_{xto0^+}frac{1}{1+e^{-1/x}}=
        frac{1}{1+0}=1
        $$






        share|cite|improve this answer


























          3












          3








          3






          You can apply l’Hôpital for the limit from the right, because it is in the form $infty/infty$.



          The limit from the left is instead in the form $0/1$, which is not indeterminate. In order to use that form of l'Hôpital for a two sided limit, you need $infty/infty$ on both sides.



          More generally you need $infty/infty$ or $0/0$ on either side.



          If you blindly apply l’Hôpital, you end up with
          $$
          lim_{xto0}frac{-e^{1/x}/x^2}{-e^{1/x}/x^2}=1
          $$
          but this would be incorrect, because
          $$
          lim_{xto0^-}frac{e^{1/x}}{e^{1/x}+1}=0
          $$
          For the limit from the right the limit is correct, but using l’Hôpital is of course not necessary:
          $$
          lim_{xto0^+}frac{e^{1/x}}{e^{1/x}+1}=lim_{xto0^+}frac{1}{1+e^{-1/x}}=
          frac{1}{1+0}=1
          $$






          share|cite|improve this answer














          You can apply l’Hôpital for the limit from the right, because it is in the form $infty/infty$.



          The limit from the left is instead in the form $0/1$, which is not indeterminate. In order to use that form of l'Hôpital for a two sided limit, you need $infty/infty$ on both sides.



          More generally you need $infty/infty$ or $0/0$ on either side.



          If you blindly apply l’Hôpital, you end up with
          $$
          lim_{xto0}frac{-e^{1/x}/x^2}{-e^{1/x}/x^2}=1
          $$
          but this would be incorrect, because
          $$
          lim_{xto0^-}frac{e^{1/x}}{e^{1/x}+1}=0
          $$
          For the limit from the right the limit is correct, but using l’Hôpital is of course not necessary:
          $$
          lim_{xto0^+}frac{e^{1/x}}{e^{1/x}+1}=lim_{xto0^+}frac{1}{1+e^{-1/x}}=
          frac{1}{1+0}=1
          $$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 4 hours ago

























          answered 4 hours ago









          egreg

          178k1484201




          178k1484201























              3














              L'Hopital's rule does not apply because the limit
              $$
              lim_{xto 0}e^{1/x}
              $$
              does not exist, since
              $$
              lim_{xto 0^-}e^{1/x}=0quadtext{while}quad lim_{xto 0^+}e^{1/x}=infty
              $$
              So, consequently
              $$
              lim_{xto 0^-}frac{e^{1/x}}{e^{1/x}+1}=0quadtext{while}quad lim_{xto 0^+}frac{e^{1/x}}{e^{1/x}+1}=1
              $$






              share|cite|improve this answer



















              • 2




                This is not the reason why l'Hôpital cannot be applied.
                – egreg
                4 hours ago
















              3














              L'Hopital's rule does not apply because the limit
              $$
              lim_{xto 0}e^{1/x}
              $$
              does not exist, since
              $$
              lim_{xto 0^-}e^{1/x}=0quadtext{while}quad lim_{xto 0^+}e^{1/x}=infty
              $$
              So, consequently
              $$
              lim_{xto 0^-}frac{e^{1/x}}{e^{1/x}+1}=0quadtext{while}quad lim_{xto 0^+}frac{e^{1/x}}{e^{1/x}+1}=1
              $$






              share|cite|improve this answer



















              • 2




                This is not the reason why l'Hôpital cannot be applied.
                – egreg
                4 hours ago














              3












              3








              3






              L'Hopital's rule does not apply because the limit
              $$
              lim_{xto 0}e^{1/x}
              $$
              does not exist, since
              $$
              lim_{xto 0^-}e^{1/x}=0quadtext{while}quad lim_{xto 0^+}e^{1/x}=infty
              $$
              So, consequently
              $$
              lim_{xto 0^-}frac{e^{1/x}}{e^{1/x}+1}=0quadtext{while}quad lim_{xto 0^+}frac{e^{1/x}}{e^{1/x}+1}=1
              $$






              share|cite|improve this answer














              L'Hopital's rule does not apply because the limit
              $$
              lim_{xto 0}e^{1/x}
              $$
              does not exist, since
              $$
              lim_{xto 0^-}e^{1/x}=0quadtext{while}quad lim_{xto 0^+}e^{1/x}=infty
              $$
              So, consequently
              $$
              lim_{xto 0^-}frac{e^{1/x}}{e^{1/x}+1}=0quadtext{while}quad lim_{xto 0^+}frac{e^{1/x}}{e^{1/x}+1}=1
              $$







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited 4 hours ago









              Szeto

              6,4362926




              6,4362926










              answered 5 hours ago









              Yiorgos S. Smyrlis

              62.6k1383163




              62.6k1383163








              • 2




                This is not the reason why l'Hôpital cannot be applied.
                – egreg
                4 hours ago














              • 2




                This is not the reason why l'Hôpital cannot be applied.
                – egreg
                4 hours ago








              2




              2




              This is not the reason why l'Hôpital cannot be applied.
              – egreg
              4 hours ago




              This is not the reason why l'Hôpital cannot be applied.
              – egreg
              4 hours ago











              0














              here, $$when~~~xto 0^+,e^{frac{1}{x}}to infty$$
              So,$$lim_{x to 0^+} frac{e^frac{1}{x}}{e^frac{1}{x}+1}=lim_{x to 0^+}frac{e^frac{1}{x}}{e^frac{1}{x}left(1+frac{1}{e^frac{1}{x}} right)}=lim_{x to 0^+}frac{1}{left(1+frac{1}{e^frac{1}{x}} right)}=frac{1}{1+0}=1$$
              $$when~~~xto 0^-,e^{frac{1}{x}} to 0$$
              So,
              $$lim_{x to 0^-} frac{e^frac{1}{x}}{e^frac{1}{x}+1}=frac{0}{0+1}=0$$



              so,simply limit doesn't exist.so we can't find the limit when $x to 0$ no matter what procedure you follow.






              share|cite|improve this answer


























                0














                here, $$when~~~xto 0^+,e^{frac{1}{x}}to infty$$
                So,$$lim_{x to 0^+} frac{e^frac{1}{x}}{e^frac{1}{x}+1}=lim_{x to 0^+}frac{e^frac{1}{x}}{e^frac{1}{x}left(1+frac{1}{e^frac{1}{x}} right)}=lim_{x to 0^+}frac{1}{left(1+frac{1}{e^frac{1}{x}} right)}=frac{1}{1+0}=1$$
                $$when~~~xto 0^-,e^{frac{1}{x}} to 0$$
                So,
                $$lim_{x to 0^-} frac{e^frac{1}{x}}{e^frac{1}{x}+1}=frac{0}{0+1}=0$$



                so,simply limit doesn't exist.so we can't find the limit when $x to 0$ no matter what procedure you follow.






                share|cite|improve this answer
























                  0












                  0








                  0






                  here, $$when~~~xto 0^+,e^{frac{1}{x}}to infty$$
                  So,$$lim_{x to 0^+} frac{e^frac{1}{x}}{e^frac{1}{x}+1}=lim_{x to 0^+}frac{e^frac{1}{x}}{e^frac{1}{x}left(1+frac{1}{e^frac{1}{x}} right)}=lim_{x to 0^+}frac{1}{left(1+frac{1}{e^frac{1}{x}} right)}=frac{1}{1+0}=1$$
                  $$when~~~xto 0^-,e^{frac{1}{x}} to 0$$
                  So,
                  $$lim_{x to 0^-} frac{e^frac{1}{x}}{e^frac{1}{x}+1}=frac{0}{0+1}=0$$



                  so,simply limit doesn't exist.so we can't find the limit when $x to 0$ no matter what procedure you follow.






                  share|cite|improve this answer












                  here, $$when~~~xto 0^+,e^{frac{1}{x}}to infty$$
                  So,$$lim_{x to 0^+} frac{e^frac{1}{x}}{e^frac{1}{x}+1}=lim_{x to 0^+}frac{e^frac{1}{x}}{e^frac{1}{x}left(1+frac{1}{e^frac{1}{x}} right)}=lim_{x to 0^+}frac{1}{left(1+frac{1}{e^frac{1}{x}} right)}=frac{1}{1+0}=1$$
                  $$when~~~xto 0^-,e^{frac{1}{x}} to 0$$
                  So,
                  $$lim_{x to 0^-} frac{e^frac{1}{x}}{e^frac{1}{x}+1}=frac{0}{0+1}=0$$



                  so,simply limit doesn't exist.so we can't find the limit when $x to 0$ no matter what procedure you follow.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 5 hours ago









                  Rakibul Islam Prince

                  952211




                  952211






























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                      When can things happen in Etherscan, such as the picture below?

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                      2 I don't know why balance of the address is 0. when can this situation? etherscan balances share | improve this question asked yesterday Ru Hi Ru Hi 11 1 New contributor Ru Hi is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct. add a comment  |  ...
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                      格利澤436b

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                      格利泽436b 太陽系外行星 太陽系外行星列表 艺术家笔下的格利泽436b 母恆星 母恆星 Gliese 436 星座 獅子座 赤經 ( α ) 11 h  42 m  11.0941 s [1] 赤緯 ( δ ) +26° 42′ 23.652″ [1] 距離 33.4 ly (10.2 pc) 光譜類型 M2.5 V [1] 軌道參數 半长轴 ( a ) 0.0291±0.0004 [2] AU 軌道離心率 ( e ) 0.150±0.012 [2] 公轉週期 ( P ) 2.643904±0.000005 [3] d (0.00723849 y) 軌道傾角 ( i ) 85.8 +0.21 −0.25 [3] ° 角距 ( θ ) 2.794 mas 近星點時間 ( T 0 ) 2,451,551.716 ±0.01 JD 半振幅 ( K ) 18.68±0.8 m/s 物理性质 质量 ( m ) 22.2±1.0 [2] M ⊕ 半徑 ( r ) 4.327±0.183 [2] [4] R ⊕ 密度 ( ρ ) 1.51 g cm -3 表面重力 ( g ) 1.18 g 溫度 ( T ) 712±36 [2] K 發現 發現時間 2004 發現者 巴特勒、沃格特、 馬西 et al. 發現方法 徑向速度、凌日法 發現地點 美國加利福尼亞州 發表論文 出版 其他名稱 Ross 905 b, GJ 436 b, [5] LTT 13213 b, GCTP 2704.10 b, LHS 310, AC+27:28217 b, Vyssotsky 616 b, HIP 57087 b, GEN# +9.80120068 b, LP 319-75 b, G 121-7 b, LSPM J1142+2642 b, 1RXS J114211.9+264328 b, ASCC 683818 b, G 147-68 b, UCAC2 41198281 b, BP...
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