# derivation of oblique asymptote for irrational functions

I understand that we can use limits to get the OA for irrational functions, but I would like to know how we were able to reach the following: $y=ax+b$---> equation of OA $a=\lim f(x)/x$ as $x$ tends to infinity and $b=\lim f(x)-ax$ as $x$ tends to infinity How did they get to those two formulas?

• What is an irrational function? – Alex M. Nov 24 '16 at 18:46
• @AlexM., one that's not rational (a ratio of two polynomials). – LSpice Nov 24 '16 at 18:49

The motivation is that, to say that the graph of $y = f(x)$ has an oblique asymptote, you want to say that the graph of $y = f(x)$ "looks like" a line for $x$ large. Hopefully it makes sense to require that the slope $f(x)/x$ of the line connecting the origin to a point $(x, y) = (x, f(x))$ on the graph of $y = f(x)$ should tend to the slope of its oblique asymptote. However, this by itself isn't enough to pin down the exact asymptote, only its slope. We need an additional piece of information, such as its $y$-intercept. If you know any point $(x, y)$ on a line of slope $a$, then you can recover the $y$-intercept of the line as $b = y - a x$. If you only know that your graph is close to a line, then you don't expect $y - a x = f(x) - a x$ to give you the exact answer, only to tend to it as $x$ becomes large; and that is why we put $b = \lim_{x \to \infty} f(x) - a x$.
Hint: an oblique asymptote is a line $y=ax+b$ to which $f(x)$ gets arbitrarily close as $x \to \infty$: $$\lim_{x \to \infty} \,f(x) - (ax + b) \, = 0$$
Dividing by $x$ gives $a = \lim_{x \to \infty} \frac{f(x)}{x}$ then replacing in the above $b = \lim_{x \to \infty} \,f(x) - ax$.