Finding $\lim_{x\to \infty}{f\left(\sqrt{\frac{2}{x}}\right)}^{x}$, when $f(0)=1$, $f^\prime(0)=0$, and $f^{\prime\prime}(0)=-1$ Let $f:\mathbb R\rightarrow \mathbb R$ such that $f^{\prime\prime}$ is continuous on $R$ and $f(0)=1$, $f^\prime(0)=0$, and $f^{\prime\prime}(0)=-1$ . Then, I have to find $$\lim_{x\to \infty}{f\left(\sqrt{\frac{2}{x}}\right)}^{x}$$
I can see that it is in indeterminate form ${1}^{\infty}$ and have let ${f\left(\sqrt{\dfrac{2}{x}}\right)}^{x}=t$ and taken $\log$ on both sides but I am not able to differentiate the function.
 A: First, note that since $f(0) > 0$ and $f$ is continuous, then $f(u)>0$ when $u$ is small enough. Thus the quantity $f\left(\sqrt{\frac{2}{x}}\right)^x$ is well-defined for $x$ large enough.
A key tool here is to use Taylor expansions, which tell you that "around $0$" (and under some regularity assumptions, which $f$ satisfies here) you can approximate $f(x)$ by the polynomial $f(0)+f'(0)x+\frac{1}{2}f''(0)x^2$. And polynomials are nice. Dealing with the case of
$
\tilde{f}(x) = 1-\frac{x^2}{2}
$
(instead of the general $f$ you have) is much easier (try is as a warmup!). This is essentially what we will be able to do.
Specifically, by Taylor's theorem (as indeed $f$ is twice differentiable at $0$) we have, using the Landau notation,
$$
f(u) = (0)+f'(0)u+\frac{1}{2}f''(0)u^2 + o(u^2)
= 1 - \frac{u^2}{2} + o(u^2)\,. \tag{1}
$$
Now, by a change of variable, letting $u\stackrel{\rm def}{=} \sqrt{\frac{2}{x}} \xrightarrow[x\to\infty]{}{0^+}$, we have
$$
f\left(\sqrt{\frac{2}{x}}\right)^x = e^{x \ln f\left(\sqrt{\frac{2}{x}}\right)}
 = e^{\frac{2}{u^2} \ln (1 - \frac{u^2}{2} + o(u^2))} \tag{2}
$$
At this point, using the standard fact that $\ln(1+u) = u+o(u)$ when $u\to0$ (to see this: it is equivalent to writing $\lim_{u\to 0}\frac{\ln(1+u)}{u}=1$), we have by (2)
$$
f\left(\sqrt{\frac{2}{x}}\right)^x 
 = e^{\frac{2}{u^2} \ln (1 - \frac{u^2}{2} + o(u^2))}
= e^{\frac{2}{u^2} (- \frac{u^2}{2} + o(u^2))}
= e^{- 1 + o(1)} \tag{3}
$$
i.e.,
$$
\lim_{x\to\infty}f\left(\sqrt{\frac{2}{x}}\right)^x 
= \lim_{u\to 0}e^{- 1 + o(1)} = \boxed{e^{-1}} \tag{4}
$$
