Find $\lim_{x\to \infty} \left[f\!\left(\sqrt{\frac{2}{x}}\,\right)\right]^x$.
\begin{align*} &\lim_{x\rightarrow\infty}\log\left(f\left(\sqrt{\dfrac{2}{x}}\right)\right)^{x}\\ &=\lim_{x\rightarrow\infty}x\log\left(f\left(\sqrt{\dfrac{2}{x}}\right)\right)\\ &=\lim_{x\rightarrow\infty}\dfrac{\log\left(f\left(\sqrt{\dfrac{2}{x}}\right)\right)}{\dfrac{1}{x}}\\ &=\lim_{x\rightarrow\infty}\dfrac{\dfrac{1}{f\left(\sqrt{\dfrac{2}{x}}\right)}\cdot f'\left(\sqrt{\dfrac{2}{x}}\right)\cdot\dfrac{1}{\sqrt{\dfrac{2}{x}}}\cdot-\dfrac{1}{x^{2}}}{-\dfrac{1}{x^{2}}}\\ &=\lim_{x\rightarrow\infty}\dfrac{\dfrac{1}{f\left(\sqrt{\dfrac{2}{x}}\right)}\cdot f'\left(\sqrt{\dfrac{2}{x}}\right)}{\sqrt{\dfrac{2}{x}}}\\ &=\lim_{x\rightarrow\infty}\dfrac{f'\left(\sqrt{\dfrac{2}{x}}\right)}{f\left(\sqrt{\dfrac{2}{x}}\right)\sqrt{\dfrac{2}{x}}}\\ &=\lim_{x\rightarrow\infty}\dfrac{f''\left(\sqrt{\dfrac{2}{x}}\right)\cdot\dfrac{1}{\sqrt{\dfrac{2}{x}}}\cdot-\dfrac{1}{x^{2}}}{\dfrac{1}{\sqrt{\dfrac{2}{x}}}\cdot-\dfrac{1}{x^{2}}\cdot f\left(\sqrt{\dfrac{2}{x}}\right)+\sqrt{\dfrac{2}{x}}\cdot f'\left(\sqrt{\dfrac{2}{x}}\right)\cdot\dfrac{1}{\sqrt{\dfrac{2}{x}}}\cdot-\dfrac{1}{x^{2}}}\\ &=\dfrac{-1}{1+0}\\ &=-1. \end{align*}
The problem is solved easily by taking logarithm. Let $2/x=t^2$ so that $t\to 0^{+}$. If $L$ is the desired limit then by continuity of logarithm $\log L$ is equal to the limit of the expression $$x\log f\left(\sqrt{\frac{2}{x}}\right)=\frac{2}{t^2}\cdot\underbrace{\frac{\log f(t)}{f(t)-1}}_{\to 1 } \cdot(f(t)-1)$$ which is equal to the limit of the expression $$\frac{f(t)-f(0)-tf'(0)}{t^2/2}$$ which is equal to $f''(0)=-1$ via Taylor's theorem (or a single application of L'Hospital's Rule). Thus $L=1/e$. Note that we only need existence of $f''(0)$ and not its continuity.
For practicing limit problems (or any math problems) this site is the best. You can ask as well as answer questions here. For a brief discussion of techniques of evaluation of limits you may refer to my blog series. As for books I always refer my favorite calculus book A Course of Pure Mathematics by G H Hardy. It should be available online for free.
This is a partial answer to your question and I'm not expecting any upvote, just to make it easier:
It seems for any polynomial with the degree equal or higher than $2$ this works fine: $f(x)=a_nx^n+a_{n-1}x^{n-1}+...-\dfrac{x^2}{2}+1$
It satisfies all conditions required by the question:
$f(0)=1, f'(0)=0, f''(0)=-1$
We may assume there are lots of function that are with this form as this can be represented by Taylor expansion of other functions when $n \to \infty$ .