Let f(x) be a polynomial satisfying $\lim_{x \to\infty} \frac{x^4f(x)}{x^8+1}=3,$ $f(2)=5 $ and $f(3)=10$, $f(-1)=2$ and $f(-6)=37$ Let f(x) be a  polynomial satisfying $\lim_{x \to\infty} \frac{x^4f(x)}{x^8+1}=3,$ $f(2)=5 $ and $f(3)=10$, $f(-1)=2$ and $f(-6)=37$ .
And then there are some question related to this.
In the solution it is given that according to the given information, $f(x)$ is a polynomial of degree 4 with leading coefficient 3.
$f(x)=3(x-2)(x-3)(x+1)(x+6)+(x^2+1)$
Now what i think is that he got the coefficient 3 using the limits and the rest of the terms$[(x-2)(x-3)(x+1)(x+6)]$ from the respective $f(x)$ values from what was given and also the $(x^2+1)$ was added for the remainders, as they satisfy all the values that have been given to us in the question. I just wanted to know if my approach is right? or is there another elegant way to do it?.

Consider the limit
$$\lim_{x \to\infty} \frac{x^4f(x)}{x^8+1}$$
$$\lim_{x \to\infty}\frac{f(x)}{x^4+\frac{1}{x^4}}$$
Since limit is finite and is equal to 3,we observe that $f(\infty) \rightarrow \infty $
Also $f^2(\infty)\rightarrow \infty$ , $f^3(\infty)\rightarrow \infty$
But after diffrentiating four times (L` Hospital rule),we see that $f^4(\infty)=72$ which implies that $ f(x)$ must be four degree polynomial
Given that $f(2)=5,f(3)=10,f(-1)=2,f(-6)=37$
Observe that output of function is input squared plus 1, so we must have $x^2+1$ involving in function
Let $f(x)=k(x-2)(x-3)(x+1)(x+6)+x^2+1$, since we don't know leading cofficient
Using, $f^4(\infty)=72$ ,we got $k=3 $
