Evaluate $limlimits_{x to 1}dfrac{x-x^x}{1-x+ln x}$.












2












$begingroup$


Problem



Evaluate $$limlimits_{x to 1}frac{x-x^x}{1-x+ln x}$$.



Solution



Consider using Taylor's formula. Expand $x^x $ and $ln x$ at $x=1$. We obtain



$$x^x=1+(x-1)+(x-1)^2+o((x-1)^2);$$
$$ln x=(x-1)-frac{1}{2}(x-1)^2+o((x-1)^2).$$
Therefore
begin{align*}
lim_{x to 1}
frac{x-x^x}{1-x+ln x}&=lim_{x to 1}frac{x-[1+(x-1)+(x-1)^2+o((x-1)^2)]}{1-x+[(x-1)-frac{1}{2}(x-1)^2+o((x-1)^2)]}\
&=lim_{x to 1}frac{-(x-1)^2-o((x-1)^2)}{-frac{1}{2}(x-1)^2+o((x-1)^2)}\
&=2.
end{align*}



Please correct me if I'm wrong! Can we solve it by L'Hospital's rule?






real-analysis limits proof-verification







share|cite|improve this question











$endgroup$












  • $begingroup$
    How do you get the expansion of $x^x$ ?
    $endgroup$
    – Yves Daoust
    Jan 16 at 14:40












  • $begingroup$
    l'Hospital's rule can be used because it is the case $0/0$ and the denominator derivative $(x-1+ln x)'neq 0$ if $xneq 1.$
    $endgroup$
    – user376343
    Jan 16 at 16:15


















2












$begingroup$


Problem



Evaluate $$limlimits_{x to 1}frac{x-x^x}{1-x+ln x}$$.



Solution



Consider using Taylor's formula. Expand $x^x $ and $ln x$ at $x=1$. We obtain



$$x^x=1+(x-1)+(x-1)^2+o((x-1)^2);$$
$$ln x=(x-1)-frac{1}{2}(x-1)^2+o((x-1)^2).$$
Therefore
begin{align*}
lim_{x to 1}
frac{x-x^x}{1-x+ln x}&=lim_{x to 1}frac{x-[1+(x-1)+(x-1)^2+o((x-1)^2)]}{1-x+[(x-1)-frac{1}{2}(x-1)^2+o((x-1)^2)]}\
&=lim_{x to 1}frac{-(x-1)^2-o((x-1)^2)}{-frac{1}{2}(x-1)^2+o((x-1)^2)}\
&=2.
end{align*}



Please correct me if I'm wrong! Can we solve it by L'Hospital's rule?






real-analysis limits proof-verification







share|cite|improve this question











$endgroup$












  • $begingroup$
    How do you get the expansion of $x^x$ ?
    $endgroup$
    – Yves Daoust
    Jan 16 at 14:40












  • $begingroup$
    l'Hospital's rule can be used because it is the case $0/0$ and the denominator derivative $(x-1+ln x)'neq 0$ if $xneq 1.$
    $endgroup$
    – user376343
    Jan 16 at 16:15
















2












2








2





$begingroup$


Problem



Evaluate $$limlimits_{x to 1}frac{x-x^x}{1-x+ln x}$$.



Solution



Consider using Taylor's formula. Expand $x^x $ and $ln x$ at $x=1$. We obtain



$$x^x=1+(x-1)+(x-1)^2+o((x-1)^2);$$
$$ln x=(x-1)-frac{1}{2}(x-1)^2+o((x-1)^2).$$
Therefore
begin{align*}
lim_{x to 1}
frac{x-x^x}{1-x+ln x}&=lim_{x to 1}frac{x-[1+(x-1)+(x-1)^2+o((x-1)^2)]}{1-x+[(x-1)-frac{1}{2}(x-1)^2+o((x-1)^2)]}\
&=lim_{x to 1}frac{-(x-1)^2-o((x-1)^2)}{-frac{1}{2}(x-1)^2+o((x-1)^2)}\
&=2.
end{align*}



Please correct me if I'm wrong! Can we solve it by L'Hospital's rule?






real-analysis limits proof-verification







share|cite|improve this question











$endgroup$




Problem



Evaluate $$limlimits_{x to 1}frac{x-x^x}{1-x+ln x}$$.



Solution



Consider using Taylor's formula. Expand $x^x $ and $ln x$ at $x=1$. We obtain



$$x^x=1+(x-1)+(x-1)^2+o((x-1)^2);$$
$$ln x=(x-1)-frac{1}{2}(x-1)^2+o((x-1)^2).$$
Therefore
begin{align*}
lim_{x to 1}
frac{x-x^x}{1-x+ln x}&=lim_{x to 1}frac{x-[1+(x-1)+(x-1)^2+o((x-1)^2)]}{1-x+[(x-1)-frac{1}{2}(x-1)^2+o((x-1)^2)]}\
&=lim_{x to 1}frac{-(x-1)^2-o((x-1)^2)}{-frac{1}{2}(x-1)^2+o((x-1)^2)}\
&=2.
end{align*}



Please correct me if I'm wrong! Can we solve it by L'Hospital's rule?






real-analysis limits proof-verification




real-analysis limits proof-verification






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 16 at 14:42









Michael Rozenberg

104k1892197




104k1892197










asked Jan 16 at 14:23









mengdie1982mengdie1982

4,897618




4,897618












  • $begingroup$
    How do you get the expansion of $x^x$ ?
    $endgroup$
    – Yves Daoust
    Jan 16 at 14:40












  • $begingroup$
    l'Hospital's rule can be used because it is the case $0/0$ and the denominator derivative $(x-1+ln x)'neq 0$ if $xneq 1.$
    $endgroup$
    – user376343
    Jan 16 at 16:15




















  • $begingroup$
    How do you get the expansion of $x^x$ ?
    $endgroup$
    – Yves Daoust
    Jan 16 at 14:40












  • $begingroup$
    l'Hospital's rule can be used because it is the case $0/0$ and the denominator derivative $(x-1+ln x)'neq 0$ if $xneq 1.$
    $endgroup$
    – user376343
    Jan 16 at 16:15


















$begingroup$
How do you get the expansion of $x^x$ ?
$endgroup$
– Yves Daoust
Jan 16 at 14:40






$begingroup$
How do you get the expansion of $x^x$ ?
$endgroup$
– Yves Daoust
Jan 16 at 14:40














$begingroup$
l'Hospital's rule can be used because it is the case $0/0$ and the denominator derivative $(x-1+ln x)'neq 0$ if $xneq 1.$
$endgroup$
– user376343
Jan 16 at 16:15






$begingroup$
l'Hospital's rule can be used because it is the case $0/0$ and the denominator derivative $(x-1+ln x)'neq 0$ if $xneq 1.$
$endgroup$
– user376343
Jan 16 at 16:15












3 Answers
3






active

oldest

votes


















2












$begingroup$

$1-x=himplies$



$$lim_{hto0}(1-h)cdotdfrac{1-(1-h)^{-h}}{h+ln(1-h)}=lim_{hto0}dfrac{1-left(1+(-h)(-h)+dfrac{(-h)(-h-1)}2cdot h^2+O(h^3)right)}{h-left(h+dfrac{h^2}2+O(h^3)right)}=dfrac1{dfrac12}$$



So, you are correct






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    Why no?
    $$lim_{xrightarrow1}frac{x-x^x}{1-x+ln{x}}=lim_{xrightarrow1}frac{1-x^x(1+ln{x})}{-1+frac{1}{x}}=lim_{xrightarrow1}frac{-x^x(1+ln{x})^2-x^xcdotfrac{1}{x}}{-frac{1}{x^2}}=2$$






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      Another solution by L'Hospital's rule



      Let $x=:1+h$. Then
      begin{align*}
      lim_{x to 1}
      frac{x-x^x}{1-x+ln x}&=lim_{h to 0}frac{(1+h)[1-(1+h)^{h}]}{ln(1+h)-h}\
      &=lim_{h to 0}frac{e^{hln(1+h)}-1}{h-ln(1+h)}\
      &=lim_{h to 0}frac{hln(1+h)}{h-ln(1+h)}\
      &=lim_{h to 0}frac{h^2}{h-ln(1+h)}\
      &=lim_{h to 0}frac{2h}{1-frac{1}{1+h}}\
      &=2lim_{h to 0}(1+h)\
      &=2.
      end{align*}






      share|cite|improve this answer









      $endgroup$













        Your Answer





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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        2












        $begingroup$

        $1-x=himplies$



        $$lim_{hto0}(1-h)cdotdfrac{1-(1-h)^{-h}}{h+ln(1-h)}=lim_{hto0}dfrac{1-left(1+(-h)(-h)+dfrac{(-h)(-h-1)}2cdot h^2+O(h^3)right)}{h-left(h+dfrac{h^2}2+O(h^3)right)}=dfrac1{dfrac12}$$



        So, you are correct






        share|cite|improve this answer









        $endgroup$


















          2












          $begingroup$

          $1-x=himplies$



          $$lim_{hto0}(1-h)cdotdfrac{1-(1-h)^{-h}}{h+ln(1-h)}=lim_{hto0}dfrac{1-left(1+(-h)(-h)+dfrac{(-h)(-h-1)}2cdot h^2+O(h^3)right)}{h-left(h+dfrac{h^2}2+O(h^3)right)}=dfrac1{dfrac12}$$



          So, you are correct






          share|cite|improve this answer









          $endgroup$
















            2












            2








            2





            $begingroup$

            $1-x=himplies$



            $$lim_{hto0}(1-h)cdotdfrac{1-(1-h)^{-h}}{h+ln(1-h)}=lim_{hto0}dfrac{1-left(1+(-h)(-h)+dfrac{(-h)(-h-1)}2cdot h^2+O(h^3)right)}{h-left(h+dfrac{h^2}2+O(h^3)right)}=dfrac1{dfrac12}$$



            So, you are correct






            share|cite|improve this answer









            $endgroup$



            $1-x=himplies$



            $$lim_{hto0}(1-h)cdotdfrac{1-(1-h)^{-h}}{h+ln(1-h)}=lim_{hto0}dfrac{1-left(1+(-h)(-h)+dfrac{(-h)(-h-1)}2cdot h^2+O(h^3)right)}{h-left(h+dfrac{h^2}2+O(h^3)right)}=dfrac1{dfrac12}$$



            So, you are correct







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 16 at 14:32









            lab bhattacharjeelab bhattacharjee

            226k15157275




            226k15157275























                2












                $begingroup$

                Why no?
                $$lim_{xrightarrow1}frac{x-x^x}{1-x+ln{x}}=lim_{xrightarrow1}frac{1-x^x(1+ln{x})}{-1+frac{1}{x}}=lim_{xrightarrow1}frac{-x^x(1+ln{x})^2-x^xcdotfrac{1}{x}}{-frac{1}{x^2}}=2$$






                share|cite|improve this answer









                $endgroup$


















                  2












                  $begingroup$

                  Why no?
                  $$lim_{xrightarrow1}frac{x-x^x}{1-x+ln{x}}=lim_{xrightarrow1}frac{1-x^x(1+ln{x})}{-1+frac{1}{x}}=lim_{xrightarrow1}frac{-x^x(1+ln{x})^2-x^xcdotfrac{1}{x}}{-frac{1}{x^2}}=2$$






                  share|cite|improve this answer









                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    Why no?
                    $$lim_{xrightarrow1}frac{x-x^x}{1-x+ln{x}}=lim_{xrightarrow1}frac{1-x^x(1+ln{x})}{-1+frac{1}{x}}=lim_{xrightarrow1}frac{-x^x(1+ln{x})^2-x^xcdotfrac{1}{x}}{-frac{1}{x^2}}=2$$






                    share|cite|improve this answer









                    $endgroup$



                    Why no?
                    $$lim_{xrightarrow1}frac{x-x^x}{1-x+ln{x}}=lim_{xrightarrow1}frac{1-x^x(1+ln{x})}{-1+frac{1}{x}}=lim_{xrightarrow1}frac{-x^x(1+ln{x})^2-x^xcdotfrac{1}{x}}{-frac{1}{x^2}}=2$$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 16 at 14:40









                    Michael RozenbergMichael Rozenberg

                    104k1892197




                    104k1892197























                        1












                        $begingroup$

                        Another solution by L'Hospital's rule



                        Let $x=:1+h$. Then
                        begin{align*}
                        lim_{x to 1}
                        frac{x-x^x}{1-x+ln x}&=lim_{h to 0}frac{(1+h)[1-(1+h)^{h}]}{ln(1+h)-h}\
                        &=lim_{h to 0}frac{e^{hln(1+h)}-1}{h-ln(1+h)}\
                        &=lim_{h to 0}frac{hln(1+h)}{h-ln(1+h)}\
                        &=lim_{h to 0}frac{h^2}{h-ln(1+h)}\
                        &=lim_{h to 0}frac{2h}{1-frac{1}{1+h}}\
                        &=2lim_{h to 0}(1+h)\
                        &=2.
                        end{align*}






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          Another solution by L'Hospital's rule



                          Let $x=:1+h$. Then
                          begin{align*}
                          lim_{x to 1}
                          frac{x-x^x}{1-x+ln x}&=lim_{h to 0}frac{(1+h)[1-(1+h)^{h}]}{ln(1+h)-h}\
                          &=lim_{h to 0}frac{e^{hln(1+h)}-1}{h-ln(1+h)}\
                          &=lim_{h to 0}frac{hln(1+h)}{h-ln(1+h)}\
                          &=lim_{h to 0}frac{h^2}{h-ln(1+h)}\
                          &=lim_{h to 0}frac{2h}{1-frac{1}{1+h}}\
                          &=2lim_{h to 0}(1+h)\
                          &=2.
                          end{align*}






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            Another solution by L'Hospital's rule



                            Let $x=:1+h$. Then
                            begin{align*}
                            lim_{x to 1}
                            frac{x-x^x}{1-x+ln x}&=lim_{h to 0}frac{(1+h)[1-(1+h)^{h}]}{ln(1+h)-h}\
                            &=lim_{h to 0}frac{e^{hln(1+h)}-1}{h-ln(1+h)}\
                            &=lim_{h to 0}frac{hln(1+h)}{h-ln(1+h)}\
                            &=lim_{h to 0}frac{h^2}{h-ln(1+h)}\
                            &=lim_{h to 0}frac{2h}{1-frac{1}{1+h}}\
                            &=2lim_{h to 0}(1+h)\
                            &=2.
                            end{align*}






                            share|cite|improve this answer









                            $endgroup$



                            Another solution by L'Hospital's rule



                            Let $x=:1+h$. Then
                            begin{align*}
                            lim_{x to 1}
                            frac{x-x^x}{1-x+ln x}&=lim_{h to 0}frac{(1+h)[1-(1+h)^{h}]}{ln(1+h)-h}\
                            &=lim_{h to 0}frac{e^{hln(1+h)}-1}{h-ln(1+h)}\
                            &=lim_{h to 0}frac{hln(1+h)}{h-ln(1+h)}\
                            &=lim_{h to 0}frac{h^2}{h-ln(1+h)}\
                            &=lim_{h to 0}frac{2h}{1-frac{1}{1+h}}\
                            &=2lim_{h to 0}(1+h)\
                            &=2.
                            end{align*}







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 16 at 14:57









                            mengdie1982mengdie1982

                            4,897618




                            4,897618






























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