Inspired from Limits: How to evaluate $\lim\limits_{x\rightarrow \infty}\sqrt[n]{x^{n}+a_{n-1}x^{n-1}+\cdots+a_{0}}-x$
What methods can be used to evaluate the limit: $$\lim_{x\rightarrow\infty} (\sqrt[p]{x^{n}+a_{n}x^{n-1}+\cdots+a_{0}}- \sqrt[q]{x^{c}+a_{t}x^{c-1}+\cdots+a_{0}})$$
PS: I am more interested in the general form where limit is of the form $\infty-\infty-\infty-...$
Hint
The idea is based on Taylor series and there's several cases to treat:
$$\sqrt[p]{x^{n}+a_{n}x^{n-1}+\cdots+a_{0}}- \sqrt[q]{x^{c}+a_{t}x^{c-1}+\cdots+a_{0}}\sim_\infty x^{n/p}-x^{c/q}$$ and the case when $\frac{n}{p}-\frac{c}{q}\ne0$ is clear and gives the limit $\pm\infty$. Now we assume that $\frac{n}{p}-\frac{c}{q}=0$ then by the Taylor series:
$$\sqrt[p]{x^{n}+a_{n}x^{n-1}+\cdots+a_{0}}- \sqrt[q]{x^{c}+a_{t}x^{c-1}+\cdots+a_{0}}\sim_\infty x^{\frac np-1}\left(\frac{a_n}{p}-\frac{a_t}{q}\right)$$
and if $\frac{a_n}{p}-\frac{a_t}{q}\ne0$ then the limit depends on three cases: $\frac np-1<0, \frac np-1>0$ or $\frac np-1=0$. If $\frac{a_n}{p}-\frac{a_t}{q}=0$ then the calculus is taken again and we push the development of the Taylor series at another term and we discus the cases that depend on $p,q,a_{n-1},a_{t-1}$ and so on.
You can divide everything under the root sign by $x^n$ and put $x$ in front of the root sign.
Then use the binomial formule of Newton for broken exponents: $$(1+y)^n=1+ny+\frac{n(n-1)}{2!}y^2+\frac{n(n-1)(n-2)}{3!}y^3+...+\frac{n(n-1)...(n-k)}{k!}y^k$$ for every real number $n$ and $y \in (-1,1)$.
