2D limit with exponentials

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)

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 $a\to0^+$. 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 $$\frac{1-xy}{(1-x)(1-y)}-\frac1{1-x}-\frac1{1-y}+1= \frac{1-xy-(1-y)-(1-x)+(1-x)(1-y)}{(1-x)(1-y)}=0.$$