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

From my multivariable textbook:

$$\lim_{|x,y|\to|0,0|}\frac{y^2\sin^2 x}{x^4+y^4}$$ (original screenshot)

Wolfram indicates that the limit DNE, but does not list the steps used to solve.

Is there a particular substitution that I'm overlooking?

share|cite|improve this question
up vote 7 down vote accepted

Calculate the limits along the lines $y=x$ and $y=2x$, as $(x,y)\to (0,0)$, these are unequal.

share|cite|improve this answer
As a procedural matter, how would one recognize the need to compare the limit when y=x against the limit when y=2x? – nitrl Mar 1 '14 at 22:06
$\frac{y^2\sin^2 x}{x^4+y^4} = \frac{x^2y^2}{x^4+y^4}\frac{\sin^2x}{x^2}$, now as $x\to 0$, we have $\frac{\sin^2x}{x^2}\to 1$. Again one writes $\frac{x^2y^2}{x^4+y^4}=\frac{1}{\frac{x^2}{y^2}+\frac{y^2}{x^2}}$, that gives the idea of trying to evaluate the limits along two lines with different slopes $y=m_1x$ and $y=m_2x$, so that the limit values become different. – r9m Mar 1 '14 at 22:13

You can look at the function on the line $y=x$. Then you look for $$lim_{x\to0}\frac{x^2\sin^2 x}{2x^4}$$. Since $sin x$ is equivalent in $0$ to $x$, you then see that this limit is $1/2$. Then look at the line $y=2x$, and the limit is now $4/17$ if i'm not mistaken. So the limit of the function in (0,0) does not exist.

share|cite|improve this answer
Thanks. I believe the second limit evaluates to 0, however. See:… – nitrl Mar 2 '14 at 22:08
You're welcome. I think it should be $$lim_{x\to0}\frac{4x^2\sin^2 x}{17x^4}$$ in your formula, so the limit is 4/17 indeed… – Asinus Mar 4 '14 at 20:19
Ah, understood. – nitrl Mar 5 '14 at 1:14

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.