2D limit with exponentials












2














I want to show that



$$lim_{(a,b)to (0,0)^+} frac{ 1-e^{-(a+b)} }{ (1-e^{-a})(1-e^{-b}) } - frac 1a - frac 1b =0 $$



where the + means that the limit is taken only with respect to paths in the nonnegative quarter .(i.e, a,b are non-negative)






real-analysis multivariable-calculus exponential-function







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    2














    I want to show that



    $$lim_{(a,b)to (0,0)^+} frac{ 1-e^{-(a+b)} }{ (1-e^{-a})(1-e^{-b}) } - frac 1a - frac 1b =0 $$



    where the + means that the limit is taken only with respect to paths in the nonnegative quarter .(i.e, a,b are non-negative)






    real-analysis multivariable-calculus exponential-function







    share|cite|improve this question

























      2












      2








      2







      I want to show that



      $$lim_{(a,b)to (0,0)^+} frac{ 1-e^{-(a+b)} }{ (1-e^{-a})(1-e^{-b}) } - frac 1a - frac 1b =0 $$



      where the + means that the limit is taken only with respect to paths in the nonnegative quarter .(i.e, a,b are non-negative)






      real-analysis multivariable-calculus exponential-function







      share|cite|improve this question













      I want to show that



      $$lim_{(a,b)to (0,0)^+} frac{ 1-e^{-(a+b)} }{ (1-e^{-a})(1-e^{-b}) } - frac 1a - frac 1b =0 $$



      where the + means that the limit is taken only with respect to paths in the nonnegative quarter .(i.e, a,b are non-negative)






      real-analysis multivariable-calculus exponential-function




      real-analysis multivariable-calculus exponential-function






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 20 '18 at 5:29









      M.A

      1599




      1599






















          1 Answer
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          I think you want $a$, $b>0$ for this to make sense. Now,
          $$frac1{1-e^{-a}}=frac{1}{a-a^2/2+cdots}=frac1a+frac12+O(a)$$
          as $ato0^+$. Therefore you can replace your limit by the limit of
          $$frac{1-e^{-a-b}}{(1-e^{-a})(1-e^{-b})}-frac1{1-e^{-a}}-frac1{1-e^{-b}}+1
          $$
          as $(a,b)to(0,0)^+$. Let $x=e^{-a}$, $y=e^{-b}$. Then this equals
          $$frac{1-xy}{(1-x)(1-y)}-frac1{1-x}-frac1{1-y}+1=
          frac{1-xy-(1-y)-(1-x)+(1-x)(1-y)}{(1-x)(1-y)}=0.$$






          share|cite|improve this answer





















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            I think you want $a$, $b>0$ for this to make sense. Now,
            $$frac1{1-e^{-a}}=frac{1}{a-a^2/2+cdots}=frac1a+frac12+O(a)$$
            as $ato0^+$. Therefore you can replace your limit by the limit of
            $$frac{1-e^{-a-b}}{(1-e^{-a})(1-e^{-b})}-frac1{1-e^{-a}}-frac1{1-e^{-b}}+1
            $$
            as $(a,b)to(0,0)^+$. Let $x=e^{-a}$, $y=e^{-b}$. Then this equals
            $$frac{1-xy}{(1-x)(1-y)}-frac1{1-x}-frac1{1-y}+1=
            frac{1-xy-(1-y)-(1-x)+(1-x)(1-y)}{(1-x)(1-y)}=0.$$






            share|cite|improve this answer


























              1














              I think you want $a$, $b>0$ for this to make sense. Now,
              $$frac1{1-e^{-a}}=frac{1}{a-a^2/2+cdots}=frac1a+frac12+O(a)$$
              as $ato0^+$. Therefore you can replace your limit by the limit of
              $$frac{1-e^{-a-b}}{(1-e^{-a})(1-e^{-b})}-frac1{1-e^{-a}}-frac1{1-e^{-b}}+1
              $$
              as $(a,b)to(0,0)^+$. Let $x=e^{-a}$, $y=e^{-b}$. Then this equals
              $$frac{1-xy}{(1-x)(1-y)}-frac1{1-x}-frac1{1-y}+1=
              frac{1-xy-(1-y)-(1-x)+(1-x)(1-y)}{(1-x)(1-y)}=0.$$






              share|cite|improve this answer
























                1












                1








                1






                I think you want $a$, $b>0$ for this to make sense. Now,
                $$frac1{1-e^{-a}}=frac{1}{a-a^2/2+cdots}=frac1a+frac12+O(a)$$
                as $ato0^+$. Therefore you can replace your limit by the limit of
                $$frac{1-e^{-a-b}}{(1-e^{-a})(1-e^{-b})}-frac1{1-e^{-a}}-frac1{1-e^{-b}}+1
                $$
                as $(a,b)to(0,0)^+$. Let $x=e^{-a}$, $y=e^{-b}$. Then this equals
                $$frac{1-xy}{(1-x)(1-y)}-frac1{1-x}-frac1{1-y}+1=
                frac{1-xy-(1-y)-(1-x)+(1-x)(1-y)}{(1-x)(1-y)}=0.$$






                share|cite|improve this answer












                I think you want $a$, $b>0$ for this to make sense. Now,
                $$frac1{1-e^{-a}}=frac{1}{a-a^2/2+cdots}=frac1a+frac12+O(a)$$
                as $ato0^+$. Therefore you can replace your limit by the limit of
                $$frac{1-e^{-a-b}}{(1-e^{-a})(1-e^{-b})}-frac1{1-e^{-a}}-frac1{1-e^{-b}}+1
                $$
                as $(a,b)to(0,0)^+$. Let $x=e^{-a}$, $y=e^{-b}$. Then this equals
                $$frac{1-xy}{(1-x)(1-y)}-frac1{1-x}-frac1{1-y}+1=
                frac{1-xy-(1-y)-(1-x)+(1-x)(1-y)}{(1-x)(1-y)}=0.$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 20 '18 at 5:42









                Lord Shark the Unknown

                101k958132




                101k958132






























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