# Problem with continuity and limits in 3 dimensions

Given the function $\lim_\limits{(x,y)\to (0,0)}$ $\frac{x^2y^2}{x^4+3y^4}$

The website I was reading lecture notes from said this function is not continuous at the point in question but doesn't go into detail about why. From what I can see if I plug in the point I get an output of $0$ and there are no problems with domain at least from what I know. How can you tell with functions of two variables whether or not a function is continuous at the given point without graphing, I am horrible at graphing in three dimensions but I am open to hearing explanations that involve graphs from those that can.

If you evaluate the limit of the function $f(x,y)$ along the curve $y=mx$, then the limit of the function comes out to be $$\lim_\limits{(x,y)\to (0,0)} \frac{x^2y^2}{x^4+3y^4}$$ $$=\lim_\limits{x\to 0} \frac{x^2(mx)^2}{x^4+3(mx)^4}$$ $$=\lim_\limits{x\to 0} \frac{m^2x^4}{x^4+3m^4x^4}$$ $$=\lim_\limits{x\to 0} \frac{m^2}{1+3m^4}$$ $$=\frac{m^2}{1+3m^4}$$ which is dependent on $m$.
So the limit of the function will be different for different $m$.
Hence the limit $\lim_\limits{(x,y)\to (0,0)} \frac{x^2y^2}{x^4+3y^4}$ does not exist and the function is not continuous at $(0,0)$.
• @K.Gibson The limit will have different values for different $m$ i.e. for different curves $y=mx$ that you wish ti calculate the limit along. But since the limit must be the same for all the curves so that the limit exists, here it is not so and hence the limit does not exist. – SchrodingersCat Jun 2 '16 at 6:34
• @K.Gibson And for your question in the first comment, NO. You apply this process only if you have to prove discontinuity. NOT for proving continuity. There you should use $\epsilon-\delta$. Works better. – SchrodingersCat Jun 2 '16 at 6:36