$\lim_{x\to\infty}\sqrt [n] {p_n(x)}-\sqrt [m] {p_m(x})$ 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 $\deg p= n$, $\deg q=m$ .
 A: 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:
$$
\frac{a_{n-1}}{na_n} - \frac{b_{m-1}}{m b_m}
$$
A: $p(x)=x^n\left(a_n+\sum_{j=0}^{n-1}a_jx^{j-n}\right)$ 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.
