Finding limit when values of derivatives at a point are given 
Let $f:\Bbb R\to\Bbb R$ be such that $f''$ is continuous on $\Bbb R$ and $f(0)=1, f'(0)=0, f''(0)=-1$. The $\displaystyle{\lim_{x\to\infty}\left(f\left(\frac{\sqrt2}{x}\right)\right)^x}$ is ..... 

I did this using particular function $f(x)=1-\frac{x^2}2$ and got answer $\frac1e$, but how to do generally?

Hint: Exponentiate the expression and express it in a form wherein you can use L'Hopital's rule. $$ \exp \ln\Biggl(f\biggl( \dfrac{\sqrt{2}}{x} \biggr)\Biggr)^x=\exp x\ln \Biggl(f\biggl( \dfrac{\sqrt{2}}{x} \biggr)\Biggr)=\exp \dfrac{\ln\Biggl(f\biggl( \dfrac{\sqrt{2}}{x} \biggr) \Biggr)}{1/x}$$Note that now you can simply make  use of L'Hopital's rule and rules of differentiation to evaluate this limit. You shouldn't be getting $1/e$, the answer is $1$.
