If a rational function has no horizontal asymptote, does it then have to have a slant asymptote This is assuming that the function is in a fractional form where the the degree of the numerator is higher than the degree of the denominator.
 A: No, a slant asymptote occurs when the numerator has a degree exactly one unit higher than the denominator. If the difference between degrees is greater, then there cnb be only vertical asymptotes.
A: Let $f(x)=P(x)/Q(x)$ be a rational function, where $P$ and $Q$ are nonzero polynomials; then


*

*$f$ has a horizontal asymptote if and only if $\deg P\le\deg Q$

*$f$ has an oblique asymptote if and only if $\deg P=1+\deg Q$
Proof. We can write $P(x)=A(x)Q(x)+R(x)$, where $\deg R<\deg Q$ (or $R(x)$ is the zero polynomial). Thus
$$
f(x)=A(x)+\frac{R(x)}{Q(x)}
$$
Clearly
$$
\lim_{x\to\pm\infty}\frac{R(x)}{Q(x)}=0
$$
Thus, if $\deg A(x)>0$ (that is, $\deg P>\deg Q$), $\lim_{x\to\pm\infty}f(x)=\pm\infty$ (with suitable choices of sign in the limit). If $A(x)=c$ is constant, then
$$
\lim_{x\to\pm\infty}f(x)=c
$$
and so $f$ has a horizontal asymptote. This is the same as saying that $\deg P\le\deg Q$.
Suppose $\deg P=1+\deg Q$; then $A(x)=ax+b$, with $a\ne0$. Then
$$
\lim_{x\to\pm\infty}\frac{f(x)}{x}=a
$$
and
$$
\lim_{x\to\pm\infty}(f(x)-ax)=b
$$
If $\deg P>1+\deg Q$, then
$$
\lim_{x\to\pm\infty}\frac{f(x)}{x}=\pm\infty
$$
(again with suitable choices of signs in the limit) and $f$ has no oblique asymptote.
