$\lim_{x\rightarrow\infty} (\sqrt[p]{x^{n}+a_{n}x^{n-1}+\cdots+a_{0}}- \sqrt[q]{x^{m}+a_{m}x^{m-1}+\cdots+a_{0}})=?$ we know that : Inspired from Limits: How to evaluate $\lim\limits_{x\rightarrow \infty}\sqrt[n]{x^{n}+a_{n-1}x^{n-1}+\cdots+a_{0}}-x$
Now :if : 
$$\lim_{x\rightarrow\infty} (\sqrt[p]{x^{n}+a_{n}x^{n-1}+\cdots+a_{0}}- \sqrt[q]{x^{m}+a_{m}x^{m-1}+\cdots+a_{0}})\to \infty- \infty$$
$$p,q,n,m \in \mathbb{N}$$
What results :$$\lim_{x\rightarrow\infty} (\sqrt[p]{x^{n}+a_{n}x^{n-1}+\cdots+a_{0}}- \sqrt[q]{x^{m}+a_{m}x^{m-1}+\cdots+a_{0}})=?$$
 A: You can do a Taylor expansion of both terms, as in this answer to the question you link:
$$
\left(x^n+a_n x^{n-1}+\dots+a_0\right)^{\frac{1}{p}}
= x^{\frac{n}{p}}\left(1+\frac{a_n}{x}+\dots+\frac{a_0}{x^n}\right)^{\frac{1}{p}}
= x^{\frac{n}{p}}\left(1+\frac{a_n}{p x}+o\left(\frac{1}{x}\right)\right)
= x^{\frac{n}{p}}+\frac{a_n}{p}x^{\frac{n}{p}-1}+o\left(x^{\frac{n}{p}-1}\right)
$$
when $x\to\infty$, so that the difference of the two terms will be
$$
x^{\frac{n}{p}}-x^{\frac{m}{q}}+\frac{a_n}{p}x^{\frac{n}{p}-1}-\frac{a_m}{q}x^{\frac{m}{q}-1}+o\left(x^{\frac{n}{p}-1}\right)+o\left(x^{\frac{m}{q}-1}\right)
$$
In particular:


*

*if $\frac{n}{p} > \frac{m}{q}$, then this becomes
$$
x^{\frac{n}{p}} + o\left(x^{\frac{n}{p}}\right) \xrightarrow[n\to\infty]{}\infty
$$

*if $\frac{n}{p} < \frac{m}{q}$, then this becomes
$$
-x^{\frac{m}{q}} + o\left(x^{\frac{m}{q}}\right) \xrightarrow[n\to\infty]{}-\infty
$$

*if $\frac{n}{p} = \frac{m}{q} < 1$, then this becomes
$$
\left(\frac{a_n}{p}-\frac{a_m}{q}\right)x^{\frac{n}{p}-1}+o\left(x^{\frac{n}{p}-1}\right)\xrightarrow[n\to\infty]{} 0
$$
as $\frac{n}{p}-1<0$.

*if $\frac{n}{p} = \frac{m}{q} > 1$ and $\frac{a_n}{p}\neq\frac{a_m}{q}$, then this becomes
$$
\left(\frac{a_n}{p}-\frac{a_m}{q}\right)x^{\frac{n}{p}-1}+o\left(x^{\frac{n}{p}-1}\right)\xrightarrow[n\to\infty]{} \begin{cases}\infty & \frac{a_n}{p}>\frac{a_m}{q}\\-\infty & \frac{a_n}{p}<\frac{a_m}{q}\end{cases}
$$
etc.

