Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$$\lim_{(x,y) \to (0,0)} \frac{{x{y^2}}}{{{x^2} + {y^4}}} = 0$$ Please, Anyone could suggest me some way for this?. Thanks.

share|cite|improve this question
Try along $x=y^2$. – AJ Stas Aug 17 '14 at 13:39

6 Answers 6

up vote 10 down vote accepted

If indeed the limit was zero then every way we approach $(0,0)$ the limit would have to be $0$.

However, if we take the path $x=y^2$ we have:

$$\lim_{x\to 0}\frac{x^2}{x^2+x^2}=\frac12\neq 0$$

So the limit cannot be zero. Maybe it could be something else, but then it would have to be $\frac12$. Take $y=0$, we have:

$$\lim_{x\to 0}\frac0{x^2}=0\neq\frac12$$

Therefore the limit does not exist.

share|cite|improve this answer
Great! many thanks. Regards. – mathsalomon Jul 23 '12 at 11:45

If $y=x$ the limit is $0$, but if $x=y^2$ the limit is $\frac{1}{2}$, then the limit don't exist.

share|cite|improve this answer

This is not true. Consider sequence $(x_n,y_n)=(n^{-2},n^{-1})$ then you get $$ \lim\limits_{n\to\infty}(x_n,y_n)=0\\ \lim\limits_{n\to\infty}\frac{x_n y_n^2}{x_n^{2}+y_n^{4}}=1 $$ If you consider another sequence $(x_n,y_n)=(n^{-1},n^{-1})$ then you get $$ \lim\limits_{n\to\infty}(x_n,y_n)=0\\ \lim\limits_{n\to\infty}\frac{x_n y_n^2}{x_n^{2}+y_n^{4}}=+\infty $$ So we conclude that limit $$ \lim\limits_{(x,y)\to(0,0)}\frac{xy^2}{x^2+y^4} $$ even doesn't exist, not to mention it is equal to $0$.

share|cite|improve this answer
Yes, I had not noticed. – mathsalomon Jul 23 '12 at 11:46

Let $x=y^2$ therefore we have $$\lim\limits_{(x,y)\to(0,0)}\frac{xy^2}{x^2+y^4}=\lim\limits_{y\to0}\frac{y^4}{2y^4}=\frac{1}{2} \ \ \ (1)$$ and if we consider $x=y$ then $$\lim\limits_{(x,y)\to(0,0)}\frac{xy^2}{x^2+y^4}=0 \ \ \ \ (2)$$ so that from (1) and (2) we see that limit don't exits

share|cite|improve this answer

Let $f(x,y):=\frac{xy^2}{x^2+y^4}$. We have $f(x^2,x)=1/2$ and $f(0,y)=0$, which proves that the limit doesn't exist.

share|cite|improve this answer

come along any $x=my^2$ then $\displaystyle \lim\limits_{(x,y)\to(0,0)}\frac{xy^2}{x^2+y^4}=\frac{1}{m^2+1}$, so for different $m$ you'll get different limits, so limit does not exists!

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.