Doubly differentiated limit

From the given derivatives at $x=0$ , the Taylor series of $f(x)$ is $f(x)=1-\frac12 x^2+\cdots$

When $x$ is large, $f\left(\sqrt{\frac{2}{x}}\right) \approx 1-\frac{1}{x}$

so the limit we want is $\lim_{x\to \infty} \left(1-\frac{1}{x}\right)^x=\boxed{e^{-1}}$

Chris Lewis gave the best possible solution. Here's my approach: $$ Let $L$ denote the value of this limit, Since the limit is in the indeterminate form $1^\infty$ , I'm tempted to use L'Hôpital's rule (LH), $\newcommand{\limxinfty}{\lim \limits_{x\to\infty} } \newcommand{\limyzero}{ \lim \limits_{y\to0^+} } \newcommand{\limzzero}{ \lim \limits_{z\to0^+} } \begin{array} {r c l l} \ln L &=& \limxinfty x \ln \left( f \left( \sqrt{\frac 2x} \right) \right) \\ \phantom0 \\ &=& \lim \limits_{x\to\infty} \dfrac{ \ln \left( f \left( \sqrt{\frac 2x} \right) \right) } { 1/x}, &\quad \text{let }y= \frac1x \\ \phantom0 \\ &=& \limyzero \dfrac{ \ln \left( f \left( \sqrt{2y} \right) \right) } { y} , & \quad \text{ this is a } \frac00 \text{ case, apply LH} \\ \phantom0 \\ &=& \limyzero \dfrac {f'(\sqrt{2y}) \cdot 1/(\sqrt {2y}) }{f(\sqrt{2y})}, \quad \text{ let }z= \sqrt{2y} \\ \phantom0 \\ &=& \limzzero \dfrac {f'(z) \cdot 1/z }{f(z)} = \lim \limits_{z\to0^+} \dfrac {f'(z) }{z \cdot f(z)}, & \quad \text{ Another } \frac00 \text{ case. Apply LH} \\ \phantom0 \\ &=& \limzzero \dfrac {f''(z) }{z \cdot f'(z) + f(z)} = \dfrac{-1}{1 + 0} = -1 \end{array}$ The answer is $L = e^{-1} \approx \boxed{0.368} .$