$\lim_{x\to\infty}\sqrt [n] {p_n(x)}-\sqrt [m] {p_m(x})$ where $p_k(x)$ is polynomial of order k.

What is the easiest way to evaluate $$\lim_{x\to\infty}\sqrt [n] {p(x)}-\sqrt [m] {q(x})$$ where $p,q\in\mathbb{R}[x]$ with $\deg p= n$, $\deg q=m$ .

-
Arbitrary polynomials? Perhaps with fixed leading coefficient? – Berci Mar 10 '13 at 11:08
@Berci : yes, the only fix is their order. – Arjang Mar 10 '13 at 11:10
This is answering your question. – Did Mar 10 '13 at 11:30

Write $$p(x) = a_n x^n + a_{n-1} x^{n-1} + o(x^{n-1}), \qquad q(x) = b_m x^m + b_{m-1} x^{m-1} + o(x^{m-1})$$ where $o(f(x))$ stands for some expression which tends to zero if divided by $f(x)$.

Note that if $n$ or $m$ is even you should suppose that the corresponding coefficient $a_n$ or $b_m$ is positive, otherwise the $n$-root is not defined for large values of $x\to +\infty$. The opposite request should hold if $x\to -\infty$.

Recall that $$\sqrt[n]{1+y} = (1+y)^{1 \over n} = 1 + \frac y n + o(y) \qquad y\to 0$$ hence for $x\to \pm\infty$ ($y=1/x\to 0$) $$\sqrt[n]{p(x)} = \sqrt[n]{a_n x^n + a_{n-1} x^{n-1} + o(x^{n-1})} = \sqrt[n]{a_n} x \sqrt[n]{1+\frac{a_{n-1}}{{a_n}x} +o(x^{-1})} = \sqrt[n]{a_n} x + \frac{a_{n-1}}{na_n} + o(1).$$ So $$\sqrt[n]{p(x)} - \sqrt[m]{q(x)} = \left(\sqrt[n]{a_n} - \sqrt[m]{b_m}\right)x + \frac{a_{n-1}}{na_n} - \frac{b_{m-1}}{m b_m} + o(1)$$ and the limit is $\pm \infty$ if the coefficient $\sqrt[n]{a_n} - \sqrt[m]{b_m}$ is different from zero (with the sign given by the sign of such coefficient), otherwise the limit is the second coefficient: $$\frac{a_{n-1}}{na_n} - \frac{b_{m-1}}{m b_m}$$

-
This is fixed leading coefficients $1$. If the leading coefficients don't match, the limit will not exist. – Berci Mar 10 '13 at 11:40
yes, the other answers show that part. which is easier... – Emanuele Paolini Mar 10 '13 at 11:42

$p(x)=x^n\left(a_n+\sum_{j=0}^{n-1}a_jx^{j-n}\right)$ and $q(x)=x^m\left(b_m+\sum_{j=0}^{m-1}b_jx^{j-m}\right)$, hence $$\sqrt[n]{p(x)}-\sqrt[m]{q(x)}=x\left(\sqrt[n]{a_n+\sum_{j=0}^{n-1}a_jx^{j-n}}-\sqrt[m]{b_m+\sum_{j=0}^{m-1}b_jx^{j-m}}\right)=:xf(x).$$ We have $\lim_{x\to +\infty}f(x)=\sqrt[n]{a_n}-\sqrt[m]{b_m}$, so if $\sqrt[n]{a_n}\neq \sqrt[m]{b_m}$ the limit is $\pm\infty$, where the signum depends whether we take the limit when $x\to \pm\infty$ and $\sqrt[n]{a_n}-\sqrt[m]{b_m}$ is positive or not.

When $\sqrt[n]{a_n}=\sqrt[m]{b_m}$, we can do a Taylor approximation of $f(x)$ in order to determine the limit.

-