Limit at infinity involving a radical difference

Multiply by the conjugate over itself: $x - \sqrt{x^2+3x} = \frac{\left(x-\sqrt{x^2+3x}\right)\left(x+\sqrt{x^2+3x}\right)}{x+\sqrt{x^2+3x}} = \frac{x^2 - (x^2+3x)}{x+\sqrt{x^2+3x}} = \frac{-3x}{x+\sqrt{x^2+3x}}.$ This removes the difference-of-radicals issue and turns the expression into a single rational-looking fraction.

Divide numerator and denominator by $x$ (valid since $x \to +\infty$ means $x>0$, so $x = \sqrt{x^2}$): $\frac{-3x}{x+\sqrt{x^2+3x}} = \frac{-3}{1+\sqrt{1+3/x}}.$ As $x \to +\infty$, $3/x \to 0$, so the denominator approaches $1+\sqrt{1} = 2$. Thus the limit is $-3/2$, meaning $y=-3/2$ is a horizontal asymptote of $h(x)$ as $x\to+\infty$.