Limit at infinity involving a radical

Multiply by the conjugate: $\sqrt{x^2+4x}-x = \frac{(\sqrt{x^2+4x}-x)(\sqrt{x^2+4x}+x)}{\sqrt{x^2+4x}+x} = \frac{4x}{\sqrt{x^2+4x}+x}.$ This converts the difference of unbounded terms into a single fraction suitable for comparing growth rates.

Divide numerator and denominator by $x$ (valid since $x>0$ as $x\to+\infty$, so $x = \sqrt{x^2}$): $\frac{4x}{\sqrt{x^2+4x}+x} = \frac{4}{\sqrt{1+\frac{4}{x}}+1}.$ As $x \to \infty$, $\frac{4}{x} \to 0$, so the expression approaches $\frac{4}{\sqrt{1}+1} = \frac{4}{2} = 2$. Thus $\lim_{x\to+\infty} h(x) = 2$, giving a horizontal asymptote $y=2$.

Apply the same conjugate technique: $\sqrt{x^2+4x}-x = \frac{4x}{\sqrt{x^2+4x}+x}.$ Now divide numerator and denominator by $x$. However, since $x < 0$ as $x \to -\infty$, we have $\sqrt{x^2} = |x| = -x$. Thus dividing by $x$ gives: $\frac{4x}{\sqrt{x^2+4x}+x} = \frac{4x}{x(\sqrt{1+\frac{4}{x}}+1)} = \frac{4}{\sqrt{1+\frac{4}{x}}+1}.$ Wait, we must be more careful. Factor $x^2$ from the radical: $\sqrt{x^2+4x} = \sqrt{x^2(1+\frac{4}{x})} = |x|\sqrt{1+\frac{4}{x}} = -x\sqrt{1+\frac{4}{x}}$ (since $x<0$, so $|x|=-x$).

Substitute back into the conjugate form: $\frac{4x}{\sqrt{x^2+4x}+x} = \frac{4x}{-x\sqrt{1+\frac{4}{x}}+x} = \frac{4x}{x(-\sqrt{1+\frac{4}{x}}+1)}.$ Divide numerator and denominator by $x$ (noting $x<0$): $\frac{4}{-\sqrt{1+\frac{4}{x}}+1} = \frac{4}{1-\sqrt{1+\frac{4}{x}}}.$ As $x \to -\infty$, $\frac{4}{x} \to 0^-$, so $1 + \frac{4}{x} \to 1^-$, meaning $\sqrt{1+\frac{4}{x}} \to 1^-$ (approaching 1 from below). Therefore $1 - \sqrt{1+\frac{4}{x}} \to 0^+$ (approaching 0 from above). Since the numerator is positive (4) and the denominator approaches $0^+$, we have $\lim_{x \to -\infty} h(x) = +\infty$. Thus $h(x)$ does not approach a finite number as $x \to -\infty$; instead it diverges to $+\infty$.