End behavior and horizontal asymptotes AB & BC

For a rational function, compare the degree $n$ of the numerator with the degree $m$ of the denominator:

1. $n < m$: the limit is $0$, so $y = 0$ is a horizontal asymptote.
2. $n = m$: the limit is the ratio of leading coefficients.
3. $n > m$: no horizontal asymptote; the outputs grow without bound.

The reasoning is one idea: divide top and bottom by the highest power of $x$ in the denominator, and every term with $x$ left underneath tends to 0. For $\frac{3x^2+1}{5x^2-x}$, dividing by $x^2$ gives $\frac{3 + 1/x^2}{5 - 1/x}$, which tends to $\frac{3}{5}$.

Applying a remembered rule without checking which case applies is the characteristic error. Dividing leading coefficients when the degrees differ produces a confident wrong answer.

A horizontal asymptote is a statement about behavior as $|x|$ grows, and says nothing about the middle of the graph. Crossing it is common and correct.

$f(x)=\frac{\sin x}{x}$ has $y = 0$ as a horizontal asymptote and crosses it infinitely often. $\frac{x}{x^2+1}$ has $y = 0$ and crosses at the origin.

This is the opposite of a vertical asymptote, which a function genuinely cannot cross, because the function is undefined there. The two are different objects and the intuition does not transfer.

Limits at $+\infty$ and $-\infty$ are separate computations and can give different answers, so a function may have two horizontal asymptotes.

The standard case involves even roots or absolute values. For $\frac{\sqrt{x^2+1}}{x}$, dividing by $x$ requires care because $\sqrt{x^2} = |x|$, which equals $x$ when $x > 0$ and $-x$ when $x < 0$. The result is $1$ as $x \to +\infty$ and $-1$ as $x \to -\infty$.

Similarly $\arctan x$ approaches $\frac{\pi}{2}$ and $-\frac{\pi}{2}$, so it has two. Computing one end and assuming the other matches is a real error whenever an even root or an absolute value is present.

Limits at infinity are not covered by direct substitution, since $\infty$ is not a number to substitute. The laws still apply to sums and products of functions that have limits at infinity, but the interesting cases are the indeterminate ones.

$\frac{\infty}{\infty}$ is indeterminate for the same reason $\frac{0}{0}$ is: the answer depends on relative growth rates, which is exactly what the degree comparison measures. Likewise $\infty - \infty$ needs work: $\sqrt{x^2+x} - x$ looks like it should be 0 but is $\frac{1}{2}$, found by multiplying by the conjugate.

So the routine mirrors Topic 1.7: recognize the form, then choose the tool. Dividing by the highest power is the standard tool at infinity, the way factoring is the standard tool at a finite point.