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.