Find $\lim_\limits{x\to 0}{(\sqrt{f^2(x)+2f(x)+3}-f(x))}$ Let $f$ be a function such that $x^2f(x)\geq x^2+x+1,\forall x\in\mathbb{R^*}$. Find the value of: $$\lim_\limits{x\to 0}{\left(\sqrt{f^2(x)+2f(x)+3}-f(x)\right)}$$
I think that we may need to apply a sandwich theorem for the limit, so let $g(x)=\sqrt{f^2(x)+2f(x)+3}-f(x)$. It is easy to prove that $g(x)>1$ near/close to 0, since $f(x)>0$ from the original relation. However, I cannot find any function $h$, such that $g(x)<h(x)$ near/close to 0 with $\lim_\limits{x\to 0}{h(x)}=1$. Any hint?

If you want to use the sandwich theorem, observe that whenever $x$ is positive,
$f(x)$ also is positive and
\begin{align}
\left(f(x) + 1 + \frac{1}{f(x)}\right)^2
 & = (f(x))^2 + 2f(x) + 3 + \frac{2}{f(x)} + \frac{1}{(f(x))^2} \\
 & > (f(x))^2 + 2f(x) + 3 \\
 & > \left(f(x) + 1\right)^2 \\
 & > 0.
\end{align}
Therefore
$$f(x) + 1 < \sqrt{(f(x))^2 + 2f(x) + 3} < f(x) + 1 + \frac{1}{f(x)}$$
and
$$1 < \sqrt{(f(x))^2 + 2f(x) + 3} - f(x) < 1 + \frac{1}{f(x)}.$$
