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Iosif Pinelis
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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.

Iosif Pinelis
  • 127.7k
  • 8
  • 107
  • 229