Quadratic function is always greater than $0$ if $$a>0$$ and $$D=0$$
Solving for $D$ we have
$$1-12a=0\Rightarrow a=\frac{1}{12}$$
So, $$a\in[\frac{1}{12},+\infty)$$
Is this the only condition to check?
For found $a\in \mathbb{A}$ how to evaluate $$\lim\limits_{x\to\infty}(x+1-\sqrt{ax^2+x+3})$$
On
We have the inequality
$$ax^2+x+3 =a\left(x+\frac{1}{2a}\right)^2+\left(3-\frac{1}{4a}\right)\ge 0$$
which is to hold for all $x\epsilon \mathscr{R}$
For $a>0$, we must have $3-\frac1{4a}\ge 0$ or
$$ a\ge \frac1{12}$$
in order for the inequality to hold for all real-valued $x$.
Else, if $a<\frac1{12}$, then $\frac1{4a}-3>0$ and for all $x<-\frac1{2a}+\sqrt{\frac{1-12a}{4a^2}}$, the inequality does not hold.
To find the limit $L=\lim_{x\to \infty}(x+1-\sqrt{ax^2+x+3})$, we note that
$$\sqrt{ax^2+x+3}=a^{1/2}x+\frac12 a^{-1/2}+O(x^{-1})$$
Thus,
$$x+1-\sqrt{ax^2+x+3}=(1-a^{1/2})x+(1-\frac12 a^{-1/2})+O(x^{-1})$$
If $a<1$, then $L=\infty$. If $a>1$, then $L=-\infty$. And if $a=1$, then $L=\frac12$.
$ax^2+x+3\ge 0$ for every $x\in\mathbb R$ if and only if $a\gt 0$ and $D=1-12a\color{red}{\le} 0$, i.e. $a\color{red}{\ge} \frac{1}{12}$.
For the limit, you can have $$\begin{align}\lim_{x\to\infty}\left(x+1-\sqrt{ax^2+x+3}\right)&=\lim_{x\to\infty}\frac{(x+1-\sqrt{ax^2+x+3})(x+1+\sqrt{ax^2+x+3})}{x+1+\sqrt{ax^2+x+3}}\\&=\lim_{x\to\infty}\frac{(x+1)^2-(ax^2+x+3)}{x+1+\sqrt{ax^2+x+3}}\\&=\lim_{x\to\infty}\frac{(1-a)x^2+x-2}{x+1+\sqrt{ax^2+x+3}}\\&=\lim_{x\to\infty}\frac{(1-a)x+1-\frac 2x}{1+\frac 1x+\sqrt{a+\frac 1x+\frac{3}{x^2}}}\end{align}$$
If $1-a\gt 0$, i.e. $a\lt 1$, then the limit is $+\infty$.
If $1-a=0$, i.e. $a=1$, then the limit is $\frac 12$.
If $1-a\lt 0$, i.e. $a\gt 1$, then the limit is $-\infty$.