End behavior and horizontal asymptote check

The numerator $4x^2+1$ has degree 2, while the denominator $2x-5$ has degree 1. Since the degree of the numerator is greater than the degree of the denominator, the function grows without bound as $x \to \pm\infty$, so we expect the limits to be infinite rather than finite.

Dividing numerator and denominator by $x$ (the highest power in the denominator) gives $h(x) = \dfrac{4x + 1/x}{2 - 5/x}$. As $x \to \infty$, this behaves like $\dfrac{4x}{2} = 2x \to \infty$, and as $x \to -\infty$, it behaves like $2x \to -\infty$. Since neither limit is a finite number $L$, $h(x)$ has no horizontal asymptote.