We have the following term:
$$ (e^{-a h}+e^{-b h})^n / 2^n$$


Now we take the limit:


$$ h\to 0, n\to \infty $$


What relation of $h$ and $n$ must be satisfied for the following limit to hold?


$$\lim\_{h\to 0, n \to \infty}(\frac{e^{-a h}+e^{-b h}}{2})^n$$
$$=\lim\_{h\to 0, n \to \infty}(1-\frac{1}{2}ah-\frac{1}{2}bh)^n $$
$$=\lim\_{h\to 0, n \to \infty}e^{-\frac{a+b}{2}h n} $$


For example, if we let $\frac{e^{-h n}}{h^2}$ keep fixed, can the above hold?



calculating a double limit
For $h\to0$, we have
$$\frac{e^{-ah}+e^{-bh}}2=1-\frac{a+b}2\,h+O(h^2)
=\exp\Big(-\frac{a+b}2\,h+O(h^2)\Big)$$
and hence
$$\Big(\frac{e^{-ah}+e^{-bh}}2\Big)^n
=\exp\Big(-\frac{a+b}2\,nh+O(nh^2)\Big).$$


So, if $h\to0$ and $nh\to c\in\mathbb R$, then $nh^2\to0$ and hence
$$\lim\Big(\frac{e^{-ah}+e^{-bh}}2\Big)^n
=\lim\exp\Big(-\frac{a+b}2\,nh\Big) \\ 
=\exp\Big(-\frac{a+b}2\,c\Big).$$




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Also, if $a>0$, $b>0$, $h\to0$ and $nh\to\infty$, then
$$-\frac{a+b}2\,nh+O(nh^2)=nh\Big(-\frac{a+b}2\,+O(h)\Big)\sim-nh \frac{a+b}2\to-\infty$$ and hence
$$\lim\Big(\frac{e^{-ah}+e^{-bh}}2\Big)^n
=\lim\exp\Big(-\frac{a+b}2\,nh\Big)=0.$$




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If $a>0$, $b>0$, $h\to0$ and $nh\to-\infty$, then similarly
$$\lim\Big(\frac{e^{-ah}+e^{-bh}}2\Big)^n
=\lim\exp\Big(-\frac{a+b}2\,nh\Big)=\infty.$$




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If $a>0$, $b>0$, $h\to0$, $n\to\infty$, but $nh$ does not converge to a limit, then
$$\lim\exp\Big(-\frac{a+b}2\,nh\Big)$$
does not exist.


