Find $\lim_{x\to \infty} \left[f\!\left(\sqrt{\frac{2}{x}}\,\right)\right]^x$. Let $f:\mathbb{R} \to \mathbb{R}$ be such that $f''$ is continuous on $\mathbb{R}$ and $f(0)=1$ ,$f'(0)=0$ and $f''(0)=-1$ . 
Then what is $\displaystyle\lim_{x\to \infty} \left[f\!\left(\sqrt{\frac{2}{x}}\,\right)\right]^x?$
When I was solving this problem, I supposed $f(x)$ to be a polynomial of degree two (because $f''$ is continuous) i.e. $f(x)=ax^2+bx+c$ and found coefficients with the help of given values . I got $f(x)=\frac{-x^2}{2} +1$. After solving , I found limit to be $e^{-1}$. I know this is a particular case.
Questions 
$1$ : Will the limit be same for all functions with these properties ? 
$2$ : Please give me some method which works for all such $f(x)$.
$3$: I want to practice more questions of this kind, please give me some references i.e. books, problem books, any online source.
Any kind of help will be highly appreciated. Thanks!

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 $n \to \infty$ .
