Let f be a scalar multivariate function : \begin{equation} f(x,y) = \begin{cases} \frac{x^2}{x^2+y^2} & \text{if } (x,y) \neq (0,0), \\ 0 & \text{if } (x,y)=(0,0). \end{cases} \end{equation}
I need to find if $f$ is continuous at $(0,0)$ or not. However, it seems not the case when looking at the $f(t,0)$ curve which isn't continous at $(0,0)$ but using the $\delta - \epsilon$ definition I find a valid $\delta$... Can you see my error in this reasoning :
$f$ is continuous at $c=(0,0)$ if and only if : $$\forall{\epsilon > 0}, \exists{\delta > 0} : ||(x,y) - (0,0)||<\delta \implies |f(x,y) - f(0,0)| <\epsilon$$ $$ \iff \sqrt{x^2 + y^2} < \delta \implies |\frac{x^2}{x^2 + y^2}| < \epsilon$$ $$ \iff \sqrt{x^2 + y^2} < \delta \implies \frac{|x|}{\sqrt{x^2 + y^2}} < \sqrt{\epsilon}$$ But since $\sqrt{x^2+y^2} < \delta$, the ($x,y)$ point lies in a ball of radius $\delta$ centered at the origin. This clearly implies that $|x| < \delta$ thus we have $\frac{|x|}{\sqrt{x^2+y^2}} < \delta$ So we just have to take $\delta = \sqrt{\epsilon}$ and the definition is guaranteed.
The error lies in the statement “This clearly implies that $|x|<\delta$ thus we have $\frac{|x|}{\sqrt{x^2+y^2}}<\delta$”. That's not true. If, say $\delta=\frac34$, $x=\frac12$ and $y=0$, then both $|x|$ and $\sqrt{x^2+y^2}$ are smaller than $\delta$, but $\frac{|x|}{\sqrt{x^2+y^2}}=1>\delta$.