# Continuity of this function as $(x,y)$ tends to $(0,0)$

Here's a function in $$x$$ and $$y$$ defined piecewise as $$f(x,y)= \left\{\begin{array}{ll} 0 & (x,y)=(0,0)\\ \frac{x^2y}{x^4+y^2} & (x,y) \neq (0,0) \\ \end{array}\right.$$ Examine its continuity as the ordered pair tends to $$(0,0)$$.

Okay, so I first tried this by assuming $$x=\frac{1}{n}$$ and $$y=\frac{1}{n^2}$$, where $$n\rightarrow \infty$$ . The limit of the function as $$(x,y) \rightarrow (0,0)$$ came out to be $$\frac{1}{2}$$, and since this is not equal to the value of the function at the said point, its discontinuous at the origin.
But when I assumed $$x=\frac{1}{n}$$ and $$y=\frac{1}{n}$$ , I got the limit zero:$$f(\frac{1}{n},\frac{1}{n})= \frac{\frac{1}{n^3}}{\frac{1}{n^4}+ \frac{1}{n^2}} = \frac{\frac{1}{n}}{\frac{1}{n^2}+ 1}$$
Since $$x,y \rightarrow 0$$,$$f(\frac{1}{n},\frac{1}{n}) \rightarrow \frac{0}{0+1}=0$$
Where am I going wrong?

• There is nothing wrong. Why can't the limits be different along different curves? Commented Feb 4, 2022 at 7:47
• With your results, decide if f is continuos at (0,0). Commented Feb 4, 2022 at 7:55
• You may use the curve $r_{\alpha}(t)=(t,\alpha t^2)$. Commented Feb 4, 2022 at 9:00
• I had just entered multivariable calculus, in fact, the very next line of my book had the answer- I believe its discontinuous because the limits are different along different paths. Its all sorted now. Thanks Commented Feb 4, 2022 at 9:23

$$\lim_{(x,y)\to(0,0)}f(x,y)$$
does not exist. If the limit does not even exist, there is no way it can equal $$f(0,0)$$, and so the function cannot be continuous at the origin.
Notice that : $$f(t, t^2) = \dfrac{t^2 t^2}{t^4 + t^2} = \dfrac{t^4}{2 t^4} = \dfrac{1}{2} \underset{t \to 0}{\to} \dfrac{1}{2} \neq f(0, 0)$$