Calculate: $limlimits_{xto0^-} frac1{ln(1-x)}+frac1x$ without LHR/Expansions [duplicate]












4












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This question already has an answer here:




  • Limit $limlimits_{xto0}frac1{ln(x+1)}-frac1x$

    4 answers





How to calculate $$limlimits_{xto0^-} left(frac1{ln(1-x)}+frac1x right)$$ without using L'Hopital, expansions nor integration?




I found the answer:



Using the Mean value theorem on:



$f(x)=e^x-frac{x^2}2-x-1$



We get:



$0lefrac{e^x-x-1}{x^2}-frac1 2le frac{e^x-x-1}{x}$



Thus:



$limlimits_{xto0^-} frac{e^x-x-1}{x^2} = frac12$



By substituting: $t=ln(1-x)$ in the original limit we get:



$limlimits_{tto0^+} frac{1-e^t+t}{t(1-e^t)} = limlimits_{tto0^+} frac{e^t-t-1}{t^2}.frac{t}{e^t-1} = frac12$






limits logarithms exponential-function







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marked as duplicate by Nosrati, max_zorn, amWhy, pre-kidney, RRL Jan 5 at 19:08


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.















  • $begingroup$
    Why "without using L'Hopital, expansions nor integration"? What is the motivation for abandoning all the convenient tools?
    $endgroup$
    – user587192
    Jan 5 at 15:52












  • $begingroup$
    Use math.stackexchange.com/questions/387333/…
    $endgroup$
    – lab bhattacharjee
    Jan 5 at 16:26
















4












$begingroup$



This question already has an answer here:




  • Limit $limlimits_{xto0}frac1{ln(x+1)}-frac1x$

    4 answers





How to calculate $$limlimits_{xto0^-} left(frac1{ln(1-x)}+frac1x right)$$ without using L'Hopital, expansions nor integration?




I found the answer:



Using the Mean value theorem on:



$f(x)=e^x-frac{x^2}2-x-1$



We get:



$0lefrac{e^x-x-1}{x^2}-frac1 2le frac{e^x-x-1}{x}$



Thus:



$limlimits_{xto0^-} frac{e^x-x-1}{x^2} = frac12$



By substituting: $t=ln(1-x)$ in the original limit we get:



$limlimits_{tto0^+} frac{1-e^t+t}{t(1-e^t)} = limlimits_{tto0^+} frac{e^t-t-1}{t^2}.frac{t}{e^t-1} = frac12$






limits logarithms exponential-function







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marked as duplicate by Nosrati, max_zorn, amWhy, pre-kidney, RRL Jan 5 at 19:08


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.















  • $begingroup$
    Why "without using L'Hopital, expansions nor integration"? What is the motivation for abandoning all the convenient tools?
    $endgroup$
    – user587192
    Jan 5 at 15:52












  • $begingroup$
    Use math.stackexchange.com/questions/387333/…
    $endgroup$
    – lab bhattacharjee
    Jan 5 at 16:26














4












4








4


1



$begingroup$



This question already has an answer here:




  • Limit $limlimits_{xto0}frac1{ln(x+1)}-frac1x$

    4 answers





How to calculate $$limlimits_{xto0^-} left(frac1{ln(1-x)}+frac1x right)$$ without using L'Hopital, expansions nor integration?




I found the answer:



Using the Mean value theorem on:



$f(x)=e^x-frac{x^2}2-x-1$



We get:



$0lefrac{e^x-x-1}{x^2}-frac1 2le frac{e^x-x-1}{x}$



Thus:



$limlimits_{xto0^-} frac{e^x-x-1}{x^2} = frac12$



By substituting: $t=ln(1-x)$ in the original limit we get:



$limlimits_{tto0^+} frac{1-e^t+t}{t(1-e^t)} = limlimits_{tto0^+} frac{e^t-t-1}{t^2}.frac{t}{e^t-1} = frac12$






limits logarithms exponential-function







share|cite|improve this question











$endgroup$





This question already has an answer here:




  • Limit $limlimits_{xto0}frac1{ln(x+1)}-frac1x$

    4 answers





How to calculate $$limlimits_{xto0^-} left(frac1{ln(1-x)}+frac1x right)$$ without using L'Hopital, expansions nor integration?




I found the answer:



Using the Mean value theorem on:



$f(x)=e^x-frac{x^2}2-x-1$



We get:



$0lefrac{e^x-x-1}{x^2}-frac1 2le frac{e^x-x-1}{x}$



Thus:



$limlimits_{xto0^-} frac{e^x-x-1}{x^2} = frac12$



By substituting: $t=ln(1-x)$ in the original limit we get:



$limlimits_{tto0^+} frac{1-e^t+t}{t(1-e^t)} = limlimits_{tto0^+} frac{e^t-t-1}{t^2}.frac{t}{e^t-1} = frac12$





This question already has an answer here:




  • Limit $limlimits_{xto0}frac1{ln(x+1)}-frac1x$

    4 answers







limits logarithms exponential-function




limits logarithms exponential-function






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edited Jan 5 at 19:12







Bilal HIMITE

















asked Jan 5 at 15:20









Bilal HIMITEBilal HIMITE

806




806




marked as duplicate by Nosrati, max_zorn, amWhy, pre-kidney, RRL Jan 5 at 19:08


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






marked as duplicate by Nosrati, max_zorn, amWhy, pre-kidney, RRL Jan 5 at 19:08


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • $begingroup$
    Why "without using L'Hopital, expansions nor integration"? What is the motivation for abandoning all the convenient tools?
    $endgroup$
    – user587192
    Jan 5 at 15:52












  • $begingroup$
    Use math.stackexchange.com/questions/387333/…
    $endgroup$
    – lab bhattacharjee
    Jan 5 at 16:26


















  • $begingroup$
    Why "without using L'Hopital, expansions nor integration"? What is the motivation for abandoning all the convenient tools?
    $endgroup$
    – user587192
    Jan 5 at 15:52












  • $begingroup$
    Use math.stackexchange.com/questions/387333/…
    $endgroup$
    – lab bhattacharjee
    Jan 5 at 16:26
















$begingroup$
Why "without using L'Hopital, expansions nor integration"? What is the motivation for abandoning all the convenient tools?
$endgroup$
– user587192
Jan 5 at 15:52






$begingroup$
Why "without using L'Hopital, expansions nor integration"? What is the motivation for abandoning all the convenient tools?
$endgroup$
– user587192
Jan 5 at 15:52














$begingroup$
Use math.stackexchange.com/questions/387333/…
$endgroup$
– lab bhattacharjee
Jan 5 at 16:26




$begingroup$
Use math.stackexchange.com/questions/387333/…
$endgroup$
– lab bhattacharjee
Jan 5 at 16:26










2 Answers
2






active

oldest

votes


















2












$begingroup$

Note that
$$begin{eqnarray}
lim_{xto0^-} left(frac1{ln(1-x)}+frac1x right)&=&lim_{xto0^+}frac{x-ln(1+x)}{xln(1+x)} \&= &lim_{xto0^+}frac{x-ln(1+x)}{x^2}cdot lim_{xto0^+}frac{x}{ln(1+x)}\&=&lim_{xto0^+}frac{x-ln(1+x)}{x^2}.
end{eqnarray}$$Now, let $$g(x) =begin{cases}frac{x-ln(1+x)}{x},quad xneq 0\ 0,quad x=0end{cases}.
$$ We can see that $g$ is continuous on $[0,1]$ and differentiable on $(0,1)$. Hence by MVT, we have
$$
L=lim_{xto0^+}frac{x-ln(1+x)}{x^2}=lim_{xto0^+}frac{g(x)}{x}=lim_{cto0^+} g'(c) =lim_{cto 0^+} frac{frac{c^2}{1+c}-(c-ln(1+c))}{c^2}=1-L.
$$ This gives $L=frac{1}{2}$.



(Justification of taking the limit) Let $h(x) = ln(1+x) - x +frac{x^2}{2}$. We have $h(x) ge 0$ since $h(0) = 0$ and $h'(x) = frac{1}{1+x}-1+xge 0$. This shows $frac{g(x)}{x}lefrac{1}{2}$. By MVT, we know that for some $cin (0,x)$,
$$
frac{1}{1+x}le frac{1}{1+c}= frac{g(c)}{c}+frac{g(x)}{x} .
$$Thus we have
$$
frac{1}{1+x}-frac{1}{2}le g(x) le frac{1}{2},
$$ and
$$
lim_{xto 0^+}g(x) =frac{1}{2}.
$$






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  • $begingroup$
    The last steps are brilliant (+1)
    $endgroup$
    – TheSimpliFire
    Jan 5 at 15:52










  • $begingroup$
    It's indeed a brilliant way, but my only problem is that I think it assumes the limits exists (not infinity).
    $endgroup$
    – Bilal HIMITE
    Jan 5 at 18:33










  • $begingroup$
    Well, that is why I added the 'justication' part at the end. Please chech that.
    $endgroup$
    – Song
    Jan 5 at 18:35



















0












$begingroup$

First you have
$$frac1{ln(1-x)}+frac1x=frac{x+ln(1-x)}{xln(1-x)}$$



Now just use Taylor's formula at order $2$:
$$x+ln(1-x)=x+bigl(-x-frac{x^2}2+o(x^2)bigr)=-frac{x^2}2+o(x^2),$$
so that $;x+ln(1-x)sim_0 =-dfrac{x^2}2$. On the other hand $;ln(1-x)sim_0 -x)$, and finally
$$frac1{ln(1-x)}+frac1xsim_0frac{-cfrac{x^2}2}{-x^2} =frac12.$$






share|cite|improve this answer









$endgroup$




















    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    Note that
    $$begin{eqnarray}
    lim_{xto0^-} left(frac1{ln(1-x)}+frac1x right)&=&lim_{xto0^+}frac{x-ln(1+x)}{xln(1+x)} \&= &lim_{xto0^+}frac{x-ln(1+x)}{x^2}cdot lim_{xto0^+}frac{x}{ln(1+x)}\&=&lim_{xto0^+}frac{x-ln(1+x)}{x^2}.
    end{eqnarray}$$    Now, let $$g(x) =begin{cases}frac{x-ln(1+x)}{x},quad xneq 0\ 0,quad x=0end{cases}.
    $$     We can see that $g$ is continuous on $[0,1]$ and differentiable on $(0,1)$. Hence by MVT, we have
    $$
    L=lim_{xto0^+}frac{x-ln(1+x)}{x^2}=lim_{xto0^+}frac{g(x)}{x}=lim_{cto0^+} g'(c) =lim_{cto 0^+} frac{frac{c^2}{1+c}-(c-ln(1+c))}{c^2}=1-L.
    $$     This gives $L=frac{1}{2}$.



    (Justification of taking the limit) Let $h(x) = ln(1+x) - x +frac{x^2}{2}$. We have $h(x) ge 0$ since $h(0) = 0$ and $h'(x) = frac{1}{1+x}-1+xge 0$. This shows $frac{g(x)}{x}lefrac{1}{2}$. By MVT, we know that for some $cin (0,x)$,
    $$
    frac{1}{1+x}le frac{1}{1+c}= frac{g(c)}{c}+frac{g(x)}{x} .
    $$    Thus we have
    $$
    frac{1}{1+x}-frac{1}{2}le g(x) le frac{1}{2},
    $$     and
    $$
    lim_{xto 0^+}g(x) =frac{1}{2}.
    $$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      The last steps are brilliant (+1)
      $endgroup$
      – TheSimpliFire
      Jan 5 at 15:52










    • $begingroup$
      It's indeed a brilliant way, but my only problem is that I think it assumes the limits exists (not infinity).
      $endgroup$
      – Bilal HIMITE
      Jan 5 at 18:33










    • $begingroup$
      Well, that is why I added the 'justication' part at the end. Please chech that.
      $endgroup$
      – Song
      Jan 5 at 18:35
















    2












    $begingroup$

    Note that
    $$begin{eqnarray}
    lim_{xto0^-} left(frac1{ln(1-x)}+frac1x right)&=&lim_{xto0^+}frac{x-ln(1+x)}{xln(1+x)} \&= &lim_{xto0^+}frac{x-ln(1+x)}{x^2}cdot lim_{xto0^+}frac{x}{ln(1+x)}\&=&lim_{xto0^+}frac{x-ln(1+x)}{x^2}.
    end{eqnarray}$$    Now, let $$g(x) =begin{cases}frac{x-ln(1+x)}{x},quad xneq 0\ 0,quad x=0end{cases}.
    $$     We can see that $g$ is continuous on $[0,1]$ and differentiable on $(0,1)$. Hence by MVT, we have
    $$
    L=lim_{xto0^+}frac{x-ln(1+x)}{x^2}=lim_{xto0^+}frac{g(x)}{x}=lim_{cto0^+} g'(c) =lim_{cto 0^+} frac{frac{c^2}{1+c}-(c-ln(1+c))}{c^2}=1-L.
    $$     This gives $L=frac{1}{2}$.



    (Justification of taking the limit) Let $h(x) = ln(1+x) - x +frac{x^2}{2}$. We have $h(x) ge 0$ since $h(0) = 0$ and $h'(x) = frac{1}{1+x}-1+xge 0$. This shows $frac{g(x)}{x}lefrac{1}{2}$. By MVT, we know that for some $cin (0,x)$,
    $$
    frac{1}{1+x}le frac{1}{1+c}= frac{g(c)}{c}+frac{g(x)}{x} .
    $$    Thus we have
    $$
    frac{1}{1+x}-frac{1}{2}le g(x) le frac{1}{2},
    $$     and
    $$
    lim_{xto 0^+}g(x) =frac{1}{2}.
    $$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      The last steps are brilliant (+1)
      $endgroup$
      – TheSimpliFire
      Jan 5 at 15:52










    • $begingroup$
      It's indeed a brilliant way, but my only problem is that I think it assumes the limits exists (not infinity).
      $endgroup$
      – Bilal HIMITE
      Jan 5 at 18:33










    • $begingroup$
      Well, that is why I added the 'justication' part at the end. Please chech that.
      $endgroup$
      – Song
      Jan 5 at 18:35














    2












    2








    2





    $begingroup$

    Note that
    $$begin{eqnarray}
    lim_{xto0^-} left(frac1{ln(1-x)}+frac1x right)&=&lim_{xto0^+}frac{x-ln(1+x)}{xln(1+x)} \&= &lim_{xto0^+}frac{x-ln(1+x)}{x^2}cdot lim_{xto0^+}frac{x}{ln(1+x)}\&=&lim_{xto0^+}frac{x-ln(1+x)}{x^2}.
    end{eqnarray}$$    Now, let $$g(x) =begin{cases}frac{x-ln(1+x)}{x},quad xneq 0\ 0,quad x=0end{cases}.
    $$     We can see that $g$ is continuous on $[0,1]$ and differentiable on $(0,1)$. Hence by MVT, we have
    $$
    L=lim_{xto0^+}frac{x-ln(1+x)}{x^2}=lim_{xto0^+}frac{g(x)}{x}=lim_{cto0^+} g'(c) =lim_{cto 0^+} frac{frac{c^2}{1+c}-(c-ln(1+c))}{c^2}=1-L.
    $$     This gives $L=frac{1}{2}$.



    (Justification of taking the limit) Let $h(x) = ln(1+x) - x +frac{x^2}{2}$. We have $h(x) ge 0$ since $h(0) = 0$ and $h'(x) = frac{1}{1+x}-1+xge 0$. This shows $frac{g(x)}{x}lefrac{1}{2}$. By MVT, we know that for some $cin (0,x)$,
    $$
    frac{1}{1+x}le frac{1}{1+c}= frac{g(c)}{c}+frac{g(x)}{x} .
    $$    Thus we have
    $$
    frac{1}{1+x}-frac{1}{2}le g(x) le frac{1}{2},
    $$     and
    $$
    lim_{xto 0^+}g(x) =frac{1}{2}.
    $$






    share|cite|improve this answer











    $endgroup$



    Note that
    $$begin{eqnarray}
    lim_{xto0^-} left(frac1{ln(1-x)}+frac1x right)&=&lim_{xto0^+}frac{x-ln(1+x)}{xln(1+x)} \&= &lim_{xto0^+}frac{x-ln(1+x)}{x^2}cdot lim_{xto0^+}frac{x}{ln(1+x)}\&=&lim_{xto0^+}frac{x-ln(1+x)}{x^2}.
    end{eqnarray}$$    Now, let $$g(x) =begin{cases}frac{x-ln(1+x)}{x},quad xneq 0\ 0,quad x=0end{cases}.
    $$     We can see that $g$ is continuous on $[0,1]$ and differentiable on $(0,1)$. Hence by MVT, we have
    $$
    L=lim_{xto0^+}frac{x-ln(1+x)}{x^2}=lim_{xto0^+}frac{g(x)}{x}=lim_{cto0^+} g'(c) =lim_{cto 0^+} frac{frac{c^2}{1+c}-(c-ln(1+c))}{c^2}=1-L.
    $$     This gives $L=frac{1}{2}$.



    (Justification of taking the limit) Let $h(x) = ln(1+x) - x +frac{x^2}{2}$. We have $h(x) ge 0$ since $h(0) = 0$ and $h'(x) = frac{1}{1+x}-1+xge 0$. This shows $frac{g(x)}{x}lefrac{1}{2}$. By MVT, we know that for some $cin (0,x)$,
    $$
    frac{1}{1+x}le frac{1}{1+c}= frac{g(c)}{c}+frac{g(x)}{x} .
    $$    Thus we have
    $$
    frac{1}{1+x}-frac{1}{2}le g(x) le frac{1}{2},
    $$     and
    $$
    lim_{xto 0^+}g(x) =frac{1}{2}.
    $$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 5 at 16:11

























    answered Jan 5 at 15:51









    SongSong

    9,644627




    9,644627












    • $begingroup$
      The last steps are brilliant (+1)
      $endgroup$
      – TheSimpliFire
      Jan 5 at 15:52










    • $begingroup$
      It's indeed a brilliant way, but my only problem is that I think it assumes the limits exists (not infinity).
      $endgroup$
      – Bilal HIMITE
      Jan 5 at 18:33










    • $begingroup$
      Well, that is why I added the 'justication' part at the end. Please chech that.
      $endgroup$
      – Song
      Jan 5 at 18:35


















    • $begingroup$
      The last steps are brilliant (+1)
      $endgroup$
      – TheSimpliFire
      Jan 5 at 15:52










    • $begingroup$
      It's indeed a brilliant way, but my only problem is that I think it assumes the limits exists (not infinity).
      $endgroup$
      – Bilal HIMITE
      Jan 5 at 18:33










    • $begingroup$
      Well, that is why I added the 'justication' part at the end. Please chech that.
      $endgroup$
      – Song
      Jan 5 at 18:35
















    $begingroup$
    The last steps are brilliant (+1)
    $endgroup$
    – TheSimpliFire
    Jan 5 at 15:52




    $begingroup$
    The last steps are brilliant (+1)
    $endgroup$
    – TheSimpliFire
    Jan 5 at 15:52












    $begingroup$
    It's indeed a brilliant way, but my only problem is that I think it assumes the limits exists (not infinity).
    $endgroup$
    – Bilal HIMITE
    Jan 5 at 18:33




    $begingroup$
    It's indeed a brilliant way, but my only problem is that I think it assumes the limits exists (not infinity).
    $endgroup$
    – Bilal HIMITE
    Jan 5 at 18:33












    $begingroup$
    Well, that is why I added the 'justication' part at the end. Please chech that.
    $endgroup$
    – Song
    Jan 5 at 18:35




    $begingroup$
    Well, that is why I added the 'justication' part at the end. Please chech that.
    $endgroup$
    – Song
    Jan 5 at 18:35











    0












    $begingroup$

    First you have
    $$frac1{ln(1-x)}+frac1x=frac{x+ln(1-x)}{xln(1-x)}$$



    Now just use Taylor's formula at order $2$:
    $$x+ln(1-x)=x+bigl(-x-frac{x^2}2+o(x^2)bigr)=-frac{x^2}2+o(x^2),$$
    so that $;x+ln(1-x)sim_0 =-dfrac{x^2}2$. On the other hand $;ln(1-x)sim_0 -x)$, and finally
    $$frac1{ln(1-x)}+frac1xsim_0frac{-cfrac{x^2}2}{-x^2} =frac12.$$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      First you have
      $$frac1{ln(1-x)}+frac1x=frac{x+ln(1-x)}{xln(1-x)}$$



      Now just use Taylor's formula at order $2$:
      $$x+ln(1-x)=x+bigl(-x-frac{x^2}2+o(x^2)bigr)=-frac{x^2}2+o(x^2),$$
      so that $;x+ln(1-x)sim_0 =-dfrac{x^2}2$. On the other hand $;ln(1-x)sim_0 -x)$, and finally
      $$frac1{ln(1-x)}+frac1xsim_0frac{-cfrac{x^2}2}{-x^2} =frac12.$$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        First you have
        $$frac1{ln(1-x)}+frac1x=frac{x+ln(1-x)}{xln(1-x)}$$



        Now just use Taylor's formula at order $2$:
        $$x+ln(1-x)=x+bigl(-x-frac{x^2}2+o(x^2)bigr)=-frac{x^2}2+o(x^2),$$
        so that $;x+ln(1-x)sim_0 =-dfrac{x^2}2$. On the other hand $;ln(1-x)sim_0 -x)$, and finally
        $$frac1{ln(1-x)}+frac1xsim_0frac{-cfrac{x^2}2}{-x^2} =frac12.$$






        share|cite|improve this answer









        $endgroup$



        First you have
        $$frac1{ln(1-x)}+frac1x=frac{x+ln(1-x)}{xln(1-x)}$$



        Now just use Taylor's formula at order $2$:
        $$x+ln(1-x)=x+bigl(-x-frac{x^2}2+o(x^2)bigr)=-frac{x^2}2+o(x^2),$$
        so that $;x+ln(1-x)sim_0 =-dfrac{x^2}2$. On the other hand $;ln(1-x)sim_0 -x)$, and finally
        $$frac1{ln(1-x)}+frac1xsim_0frac{-cfrac{x^2}2}{-x^2} =frac12.$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 5 at 16:02









        BernardBernard

        119k740113




        119k740113















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            WPF add header to Image with URL pettitions [duplicate]

            Image
            up vote 0 down vote favorite This question already has an answer here: Image source URI requiring authorization 3 answers Add header to all* HTTP requests on a machine 1 answer first of all sorry for my english. Recently I added some security in the backend of my project, and now I need to send a validation code(apiKey) inside the headers to retrieve the images. I have a problem with ALL images in my project that I had attached in my UI. For example, before I had: <Image Height="90" Source="{Binding Video.Thumbnail}"/> Now I need to change my code to create a CUSTOM petition add the apiKey in the header:...
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