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's the value of $$ \lim_{(\alpha,\beta)\to(0,0)}\frac{[\lambda_i\alpha+\mu_i\beta+O(r^2)] [\lambda_j\alpha+\mu_j\beta+O(r^2)][-\lambda\alpha^2-\mu\beta^2+O(r^3)]} {\left(\sum_{i=1}^3\left(\lambda_i\alpha+\mu_i\beta+O(r^2)\right)^2\right)^{3/2}},\quad i,j = 1,2,3\tag{*} $$ where $\lambda_i,\mu_i,\lambda,\mu$ are constants, $r = \sqrt{\alpha^2+\beta^2}$?

I guess this limit may not exist, OR if it does exist, it should be $0$. For investigating the possibility, I tried some simpler case. For instance, let $\lambda_i=\mu_i=-\lambda=-\mu=1$ for $i=1,2,3$ and $O(r^3)=O(r^2)=0$. Then the limit becomes $$ \lim_{(\alpha,\beta)\to(0,0)}\frac{(\alpha+\beta)^2(\alpha^2+\beta^2)}{3^{3/2}(\alpha+\beta)^3}. $$

How should I deal with the term such as $$ \lim_{(\alpha,\beta)\to(0,0)}\frac{\alpha^4}{(\alpha+\beta)^3} ? $$ In general what's the value of (*)?

For the polynomial with only one variable, we have $$ \lim_{x\to 0}\frac{f(x)}{g(x)}=0 $$ when the order of $f$ is higher than that of $g$.

Do we have the same result for the multivariate polynomials which appears in the title?

share|cite|improve this question
If the limit exists, then approaching the origin on the line $\beta = 0$, for example, will give you the limit. This holds for any path to the origin, $\alpha=\beta$, $\alpha = 0$, etc. – Patrick Mar 18 '12 at 18:26
up vote 0 down vote accepted

In light of the answer and comments to this question, I think the polar coordinates can help. Let $$\alpha=r\cos\theta,\quad\beta=r\sin\theta$$(*) can finally be reduced to $$ \lim_{r\to 0}\frac{f(r)}{g(r)}Q(\sin\theta,\cos\theta,r) $$ where $Q$ is bounded, and $f,g$ are polynomials with respect to $r$.

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.