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 $= \sqrt{1}+0=1$?
 A: 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. 
A: Since $x$ is large $$f(x)= \sqrt{x^2+4x+3} +x= \sqrt{x^2\Big(1+\frac{4}{x}+\frac{3}{x^2}\Big)}+x=x\sqrt{1+\frac{4}{x}+\frac{3}{x^2}}+x$$ Now consider $\sqrt{1+y}$ when $y$ is small; the Taylor expansion built at $y=0$ gives $$\sqrt{1+y}=1+\frac{y}{2}-\frac{y^2}{8}+O\left(y^3\right)$$ Relace now $y$ by $(\frac{4}{x}+\frac{3}{x^2})$; so $$\sqrt{1+\frac{4}{x}+\frac{3}{x^2}}=1+\frac{2}{x}-\frac{1}{2
   x^2}+O\left(\left(\frac{1}{x}\right)^3\right)$$ So $$x\sqrt{1+\frac{4}{x}+\frac{3}{x^2}}+x \simeq 2x+2-\frac{1}{2
   x}$$ and then $f(x$ behaves just as $(2x+2)$ which is an oblique asymptote for the curve which stays below its asymptote
