$f(x) = \begin{cases} \frac{xy^2}{x^2+y^4}, (x,y) \ne0 \\0, (x,y) = 0 \end{cases}$ Determine if the function is continuous at the origin. Let $f: \mathbb{R^2} \to \mathbb{R}$, $$f(x) = \begin{cases} \frac{xy^2}{x^2+y^4}, (x,y) \ne0 \\0, (x,y) = 0 \end{cases}$$ Determine if the function is continuous at the origin.

Now when $(x,y) \ne 0$, letting $x = y^2$ I have that $\frac{y^4}{y^4+y^4} = \frac{1}{2}$. Similarly when $y=x$ I get $\frac{x^3}{x^2+x^4} = \frac{x}{1+x^2} \to 0$, hence the function is not continuous at the origin. Now I tried using epsilon-delta for this, but couldn't get it to work. What's would be wrong with my approach?
We have that $|f(x)-f(0)| &lt; \varepsilon$ whenever $\|x-0\| = \|x\| = |x| &lt; \delta$.
So let $\varepsilon &gt;0$ and $\delta= \frac{\varepsilon}{y^2}$ $$|f(x)-f(0)| = |f(x)| = \frac{|x|y^2}{|x^2+y^4|} \leqslant |x|y^2 &lt; \delta y^2 =\varepsilon.$$
But this is not true since just above we showed that the function is not continuous at the origin. I guess the problem is $\frac{|x|y^2}{|x^2+y^4|} \leqslant |x|y^2$ which isn't probably true?
The statement &quot;$f$ is continuous at $(0,0)$&quot; has the definition: &quot;for every $\varepsilon&gt;0$, there exists a $\delta&gt;0$ such that for every $(x,y)\in\Bbb R^2$, if $(x,y)$ is within $\delta$ of $(0,0)$, then $f(x,y)$ is within $\varepsilon$ of $f(0,0)$&quot;.
The first quantified variable is $\varepsilon$, which must be an arbitrary constant (not depending on any other variables).
The next quantified variable is $\delta$, which can be chosen in terms of $\varepsilon$ (since that variable has already been quantified), but must be independent of $x$ and $y$ (since those variables have not been quantified yet when we get to $\delta$). This is the source of the error in the proof—you set $\delta= \varepsilon/y$ which is not allowed.
(And the next quantified variables are $x$ and $y$, which are allowed to be constrained by $\delta$ since that variable has already been quantified.)
The statement &quot;$f$ is continuous at $(0,0)$&quot; has the definition: &quot;for every $\varepsilon&gt;0$, there exists a $\delta&gt;0$ such that for every $(x,y)\in\Bbb R^2$, if $(x,y)$ is within $\delta$ of $(0,0)$, then $f(x,y)$ is within $\varepsilon$ of $f(0,0)$&quot;.
The first quantified variable is $\varepsilon$, which must be an arbitrary constant (not depending on any other variables).
The next quantified variable is $\delta$, which can be chosen in terms of $\varepsilon$ (since that variable has already been quantified), but must be independent of $x$ and $y$ (since those variables have not been quantified yet when we get to $\delta$). This is the source of the error in the proof—you set $\delta= \varepsilon/y$ which is not allowed.
(And the next quantified variables are $x$ and $y$, which are allowed to be constrained by $\delta$ since that variable has already been quantified.)