We usually say a funciton has an oblique asymptote $y=ax+b$ if
$\lim_{x\to \infty}\frac{f(x)}{x}=a \text{ ; and} $ $\lim_{x\to \infty}{f(x)}-{ax}=b $
This is basically saying, for $x$ large enough, the function resembles the line $y=ax+b$. This is a special case of what we call "asymptotic behaviour" of a function. Particularily, given two polynomial functions $p$ and $q$, the quotient
$f(x)=\frac{p(x)}{q(x)}$
will behave asymptotically like a line $y=ax+b$ if $\deg p =\deg q+1$. Why is this? Let's examine the polynomials
$q(x)=a_nx^n+a_{n-1}x^{n-1}\cdots+a_0 \text{ ; } a_n\neq0$
$p(x)=b_{n+1}x^{n+1}+b_{n}x^{n}\cdots+b_0 \text{ ; } b_{n+1}\neq0$
We have that their quotient is
$f(x)=\frac{{p(x)}}{{q\left( x \right)}} = \frac{{{b_{n + 1}}{x^{n + 1}} + {b_n}{x^n} \cdots + {b_0}{\text{ }}}}{{{a_n}{x^n} + {a_{n - 1}}{x^{n - 1}} \cdots + {a_0}}}$
And now, if we divide by $x$ and find the limit for $x\to \infty$, we find
$\mathop {\lim }\limits_{x \to \infty } \frac{{f\left( x \right)}}{x} = \mathop {\lim }\limits_{x \to \infty } \frac{1}{x}\frac{{{b_{n + 1}}{x^{n + 1}} + {b_n}{x^n} \cdots + {b_0}{\text{ }}}}{{{a_n}{x^n} + {a_{n - 1}}{x^{n - 1}} \cdots + {a_0}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{{b_{n + 1}}{x^n} + {b_n}{x^{n - 1}} \cdots + \frac{{{b_0}}}{x}{\text{ }}}}{{{a_n}{x^n} + {a_{n - 1}}{x^{n - 1}} \cdots + {a_0}}}$
Note that now, except for the insignificant $\dfrac{b_0}{x}$ we have a quotient of polynomials of equal degree, precisely because upon dividing by $x$ the degrees were equated. So this limit will be constant, in fact we have
$\mathop {\lim }\limits_{x \to \infty } \frac{{f\left( x \right)}}{x} = \mathop {\lim }\limits_{x \to \infty } \frac{{{b_{n + 1}}{x^n} + {b_n}{x^{n - 1}} \cdots + \frac{{{b_0}}}{x}{\text{ }}}}{{{a_n}{x^n} + {a_{n - 1}}{x^{n - 1}} \cdots + {a_0}}} = \frac{{{b_{n + 1}}}}{{{a_n}}}$
Calculating the second part of the limit, we get
$\mathop {\lim }\limits_{x \to \infty } f\left( x \right) - \frac{{{b_{n + 1}}}}{{{a_n}}}x = \mathop {\lim }\limits_{x \to \infty } \frac{{{b_{n + 1}}{x^{n + 1}} + {b_n}{x^n} \cdots + {b_0}{\text{ }}}}{{{a_n}{x^n} + {a_{n - 1}}{x^{n - 1}} \cdots + {a_0}}} - \frac{{{b_{n + 1}}}}{{{a_n}}}x$
because of the factor of $\dfrac{{{b_{n + 1}}}}{{{a_n}}}$ the $x^{n+1}$ will cancel you (the algebra of that is a bit messy) but you'll end up again with a quotient of polynomials of equal degree:
$\mathop {\lim }\limits_{x \to \infty } \frac{{\left( {{b_n} - \frac{{{b_{n + 1}}}}{{{a_n}}}{a_{n - 1}}} \right){x^n} + \cdots + {b_0}{\text{ }}}}{{{a_n}{x^n} + {a_{n - 1}}{x^{n - 1}} \cdots + {a_0}}} = \frac{{{b_n}}}{{{a_n}}} - \frac{{{b_{n + 1}}}}{{{a_n}}}\frac{{{a_{n - 1}}}}{{{a_n}}}$
so the asymptote you'll get is
$y = \frac{{{b_{n + 1}}}}{{{a_n}}}x + \frac{{{b_n}}}{{{a_n}}} - \frac{{{b_{n + 1}}}}{{{a_n}}}\frac{{{a_{n - 1}}}}{{{a_n}}}$
And why do we get a straight line only when the difference is $1$? That because we divide by $x^1$ (which lowers the degree of the denominator by $1$). However, if the degrees differs by say $r$, the rational function will behave asymptotically like a polynomial of degree $r$