$\lim_{x\rightarrow\infty} (\sqrt[p]{x^{n}+a_{n}x^{n-1}+\cdots+a_{0}}- \sqrt[q]{x^{m}+a_{m}x^{m-1}+\cdots+a_{0}})=?$

Question

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 :$$\lim_{x\rightarrow\infty} (\sqrt[p]{x^{n}+a_{n}x^{n-1}+\cdots+a_{0}}- \sqrt[q]{x^{m}+a_{m}x^{m-1}+\cdots+a_{0}})=?$$

You could do a Taylor expansion of both, to see what is going on -- as in this answer of the question you linked. — Clement C., Jan 19 '17 at 15:18
Please see my answer to this question http://math.stackexchange.com/questions/2104344/compute-lim-limits-n-to-infty-frac-sqrt4n44n-sqrt3n33n/2104431#2104431 also just today strangely. — Rene Schipperus, Jan 19 '17 at 15:39
After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark ✓ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?. — Clement C., Jan 20 '17 at 19:48

score 2 · Accepted Answer · edited Apr 13 '17 at 12:21

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.

$\lim_{x\rightarrow\infty} (\sqrt[p]{x^{n}+a_{n}x^{n-1}+\cdots+a_{0}}- \sqrt[q]{x^{m}+a_{m}x^{m-1}+\cdots+a_{0}})=?$

1 Answers1

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