Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\deg p= n$, $\deg q=m$ .

share|cite|improve this question
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
up vote 2 down vote accepted

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: $$ \frac{a_{n-1}}{na_n} - \frac{b_{m-1}}{m b_m} $$

share|cite|improve this answer
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.