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

I'm pretty sure that

\begin{equation} \lim_{(x,y) \rightarrow (0,0)} \frac{x^4y}{x^2 + y^2} = 0, \end{equation}

but I'm having some trouble proving it.

The only technique I'm aware of that can be used to show indeterminate limits of $\geq 2$ variables exist is the Squeeze Theorem. I've tried applying it here (by assuming $|y| < 1$ and bounding the quantity of interest by $\pm\frac{x^4y}{x^2 + y^2}$), but I didn't get anywhere.

Any help is appreciated.

share|cite|improve this question
up vote 9 down vote accepted

Since $|xy|\leq \frac{x^2+y^2}{2}$ by the GM-QM inequality, you simply have: $$\left|\frac{x^4 y}{x^2+y^2}\right|\leq \frac{1}{2}|x|^3.$$

share|cite|improve this answer

$$ 0\le\left|\frac{x^4y}{x^2 + y^2}\right|= \left|\frac{x^4y}{\|(x,y)\|^2}\right|\le \frac{\|(x,y)\|^5}{\|(x,y)\|^2}= \|(x,y)\|^3. $$

share|cite|improve this answer

Use polar coordinates: $x = r\cos t$, $y = r\sin t\Rightarrow x^2 + y^2 = r^2$, and $yx^4 = r^5\sin t(\cos t)^4$. So: $yx^4/(x^2 + y^2) = r^3\sin t(\cos t)^4 \to 0$ as $r\to 0$.

share|cite|improve this answer

$$ \left|\frac{x^4y}{x^2+y^2}\right| \le \left|\frac{x^4y}{x^2}\right| = \left|x^2y\right| \to 0. $$

share|cite|improve this answer

The basic inequality here is $x^2\le x^2+y^2$, from which you get $$ \frac{|x|}{\sqrt{x^2+y^2}}\le 1 $$ with a similar one for $y$. So $$ \left|\frac{x^4y}{x^2+y^2}\right|\le|x^3|. $$

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.