The limit of $f(x)= \sqrt{x^2+4x+3} +x$ as $x\to\infty$ The problem is to find $\lim_{x\to\infty} \sqrt{x^2+4x+3} +x$.
Do I just divide everything by $x^2$ and get limit $= \sqrt{1}+0=1$?

According to what was given, with the assumption that the function $f: x \mapsto \sqrt{x^{2} + 4x + 3} + x$ is defined on $\mathbb{R}$, it is concluded that for every $\varepsilon > 0$ there is a real $X$ such that if $x \geq X$ then 
$$|\sqrt{x^{2} + 4x + 3} + x - l| \geq \varepsilon,$$
viz, the function in question grows indefinitely as $x$ goes beyond every bound.
For, given any $\varepsilon > 0$ we can choose a real $X$ such that
$$|\sqrt{X^{2} + 4X + 3} + X| \geq \varepsilon,$$
so that for all real $x \geq X$ we have
$$|\sqrt{x^{2} + 4x + 3} + x| \geq \varepsilon$$
that follows from the monotonicity of the function in question. 
