Finding asymptote of $\sqrt[n]{p(x)}$

Question

Suppose $p(x) = ax^n+bx^{n-1} + cx^{n-2}+ \dots + z $ . Now find asymptote of $\sqrt[n]{p(x)}$ . I looked for it but I didn't find it and also couldn't solve it .

Example :

Answer 1

Perhaps it helps to notice that

$$\sqrt[n]{a_n x^n+\dots+a_0} = x\sqrt[n]{a_n+\frac{a_{n-1}}{x}\dots+\frac{a_{0}}{x^n}}$$

if $x > 0$.

Answer 2

In the same spirit as Argon's answer $$p(x)=\sum_{i=0}^n a_ix^i=a_nx^n\left(1+ \sum_{i=0}^{n-1} \frac {a_i}{a_n x^{n-i}}\right)$$ Assume (wlog) $a_n >0$

$$\sqrt[n]{p(x)}=x\sqrt[n]{a_n}\left(1+ \sum_{i=0}^{n-1} \frac {a_i}{a_n x^{n-i}}\right)^{1/n}$$ Now, use the generalized binomial theorem (or Taylor series) to get $$\sqrt[n]{p(x)}=x\sqrt[n]{a_n}\left(1+\frac{a_{n-1}}{na_n x}+\cdots\right)=x\sqrt[n]{a_n}+\frac{a_{n-1}\sqrt[n]{a_n}}{n\,a_n}\cdots$$ which is the asymptote.