When we find oblique asymptotes, we divide the numerator by the denominator and take only the polynomial portion of the expression as the equation of the slant asymptote. Why? What is the proof that this equation is the slant asymptote?
[Math] Oblique asymptotes
calculus
Best Answer
Because you're interested in the behaviour of the function as $|x|$ grows large. You can split the function up. I'll take an example:
$$\frac{3x^2+1}{x+2}$$ $$\frac{3x(x+2)-6x+1}{x+2}$$ $$3x+\frac{-6x+1}{x+2}$$ $$3x+\frac{-6(x+2)+13}{x+2}$$ $$3x-6+\frac{13}{x+2}$$
at each step I'm rewriting the highest term in a way that involves $x+2$, and then adjusting the lower terms so that the expansion comes out right. Now look what happens to the fraction as $x$ grows large:
$$\frac{13}{102},\frac{13}{1002},\frac{13}{10002}, \cdots$$
It gets smaller and smaller and goes to $0$. You can show the same thing for the other direction as well (negative numbers that get closer and closer to zero). So an asymptote is concerned with behaviour as $x$ grows large, and as $x$ grows large the fractional portion drops out to $0$. It only influences the function near the origin, which the asymptote isn't concerned with.