ISI 2016 B.Math Entrance (3)

Note that

$g(1) = \displaystyle \lim_{x \to \infty} \dfrac{f(x)}{x^3} = 1$

Now

$\displaystyle \lim_{x \to \infty} \dfrac{f(x)}{x^3} = \begin{cases} 0 & \text{if degree } f(x) < 3 \\ a & \text{if degree } f(x) = 3 \\ \infty & \text{if degree } f(x) > 3 \end{cases}$

where $a$ is the leading coefficient of the polynomial $f(x)$ when it has a degree 3.

Thus, we find that $f(x)$ should be a monic cubic polynomial to satisfy the condition $g(1) = 1$ .

Let

$f(x) = x^3 + bx^2 + cx + d \\ \implies f(xy) = x^3 y^3 + b x^2 y^2 + c xy + d$

Hence

$g(y) = \displaystyle \lim_{x \to \infty} \dfrac{f(xy)}{x^3} = \displaystyle \lim_{x \to \infty} \left( y^3 + \dfrac{b y^2}{x} + \dfrac{cy}{x^2} + \dfrac{d}{x^3} \right) = \boxed{y^3}$

Relevant wiki: Big O Notation

From $\lim_{x \to \infty} \frac{f(x)}{x^3} = 1$ we obtain that $f(x) \sim x^3$ . Or what is equivalent, that $f(x) = x^3 + o(x^3)$ . Using this:

$g(y) = \lim_{x \to \infty} \frac{f(xy)}{x^3} = \lim_{x \to \infty} \frac{x^3 y^3 + o(x^3 y^3)}{x^3} = \lim_{x \to \infty} y^3 + \frac{o(x^3)}{x^3} = y^3 + 0 = y^3$ .