Prove equation with Taylor, change of variable and Hopital If $f\in C^2(\mathbb{R})$  such that,
$$f(0)=1, \quad f'(0) =0,$$
and $$f''(0) = -1,$$
then $\forall a\in \mathbb{R}$ we have
$$\lim _{x\to +\infty} \left( f(\frac{a}{\sqrt{x}}) \right)^x = e^{-\frac{a^2}{2}}.$$
Is that true? And how I can prove that using taylor, a change of variable and Hopital?

By using a Taylor series expansion, as $x \to \infty$, one has
$$
f\Big(\frac{a}{\sqrt{x}}\Big) =f(0)+\frac{a}{\sqrt{x}}\cdot f'(0)+\frac{a^2}{x}\cdot \frac{f''(0)}{2!}+\mathcal{O}\left(\frac1{x^{3/2}} \right)
$$ that is
$$
f\Big(\frac{a}{\sqrt{x}}\Big) =1-\frac{a^2}{2x}+\mathcal{O}\left(\frac1{x^{3/2}} \right)
$$ giving, as $x \to \infty$,

$$
\left(f\Big(\frac{a}{\sqrt{x}}\Big)\right)^x=e^{x \log \left(1-\frac{a^2}{2x}+\mathcal{O}\left(\frac1{x^{3/2}} \right) \right)}=e^{-\frac{a^2}{2}+\mathcal{O}\left(\frac1{\sqrt{x}}\right)} \to e^{-\frac{a^2}{2}},
$$ 

since $\log(1-u)=-u+O(u^2)$ as $u \to 0.$
