Verify, using Taylor, a change of variable, the theorem of Hopital two times and WITHOUT any notable special limit, the following equation:
if $f\in C^2(\mathbb{R})$ such that $$f(0)=1, \quad f'(0) =0, \quad f''(0) =-1$$ then for every $a\in \mathbb{R}$ $$\lim _{x\to +\infty} \left( f(\frac{a}{\sqrt{x}}) \right)^x = e^{-\frac{a^2}{2}}.$$
Note: this does not use L'Hospital, but does use change of variables and Taylor (only noticed requirement to use L'Hospital after completing the answer...). Hope it still helps.
For any $t$ in some neighborhood of zero, $f(t)>0$ because $f(0)=1$ and $f$ is continuous. Therefore,
$$ f(t)^x = e^{x \ln f(t)}.$$
Now by Taylor expansion, for any $t$,
$$f(t) = 1 + f''(c)\frac{t^2}{2},$$
where $c$ is between $0$ and $t$.
Therefore for $t= \frac{a}{\sqrt{x}}$, we have
$$ x \ln f(t) = x \ln (1+f''(c) \frac{a^2}{2x})=(*)$$
By Taylor expansion, for $|z|<1$.
$$\ln (1+z)= z - \frac{z^2}{2(1+{\bar{c}})^2},$$
where $\bar{c}$ is between $0$ and $z$.
For $x$ large enough, letting $z= f''(c) \frac{a^2}{2x}$, we clearly have $|z|<1$, and so
$$ (*) = x \frac{ f''(c) a^2}{2x} + x \frac{1}{2(1+\bar{c})^2}\frac{f''(c)^2 a^4}{ 4x^2}=\frac{ f''(c) a^2}{2}+\frac{1}{2(1+\bar{c})^2}\frac{f''(c)^2 a^4}{ 8x}.$$
Finally, as $x\to \infty$, $t\to 0$, hence $c\to 0$ and also $z\to 0$, so ${\bar{c}}\to 0$. Since $f''$ is continuous and $f''(0)=-1$, we obtain
$$ \lim_{x\to\infty} (*) = -\frac{a^2}{2}.$$
$$ \lim_{x\to \infty} f(\frac{a}{\sqrt{x}})^x = \lim_{x\to \infty} e^{ (*) } =e^{\lim_{x\to \infty} (*)} = e^{-\frac{a^2}{2}}.$$
The result follows from continuity of the exponential function.