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!

Hint: the limit is an indeterminate form $[1^\infty]$. Consider its logarithm
$$
x \log \left( f \left( \sqrt{\frac{2}{x}} \right)\right),
$$
assuming that $f>0$. Now you should ask yourself how to expand the argument of the logarithm as $x \to +\infty$. Of course, you will need an expansion of $f$ near zero.
Then you can try to compare the logarithm with $x$, and hope to get a conclusion.
