TITLE: Find $\lim_{x\to \infty} \left[f\!\left(\sqrt{\frac{2}{x}}\,\right)\right]^x$.
QUESTION [7 upvotes]: 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!

REPLY [4 votes]: 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.