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