Determine if $f(x,y)=\left\{\begin{matrix} xy\cos(\frac{1}{x^2+y^4}) & (x,y) \ne (0,0)\\ 0 & (x,y)=(0,0) \end{matrix}\right.$ continuous at $(0,0)$

Question

Determine if $f(x,y)=\left\{\begin{matrix} xy\cos(\frac{1}{x^2+y^4}) & (x,y) \ne (0,0)\\ 0 & (x,y)=(0,0) \end{matrix}\right.$ is continuous at $(0,0)$

I have tried looking at different directions, polar coordinates, and the squeeze rule.

With the polar, I got a function that I don't know what to do with, and with the squeeze rule I always have $x$ and $y$.

Any help?

Thanks!

Answer 1

Let ${(x_n, y_n)}_{n \in \mathbb N}$ be a sequence in $\mathbb R^2$, such that $$ \lim_{n \to \infty} (x_n, y_n) = (0, 0) \; .$$ Note that $$ 0 \leq \left| f(x_n,y_n) \right| \leq \left| x_n y_n \right| $$ for each $n \in \mathbb N$.

By the squeeze theorem we see that $$ \lim_{n \to \infty} \left| f(x_n, y_n) \right| = 0 \; , $$ i.e. $$ \lim_{n \to \infty} f(x_n, y_n) = 0 = f(0, 0)\; , $$ so $f$ is continuous at $(0, 0)$.

Answer 2

If you want to prove that $f$ is continuous at $(x,y)=(0,0)$, so you should show that $$\lim_{(x,y)\to (0,0)}f(x,y)=f(0,0).$$ By definition on $f$, you have that $$f(0,0)=0$$ and since that $$-1\leqslant \cos\left(\frac{1}{x^{2}+y^{4}}\right)\leqslant 1$$ So, $$\lim_{(x,y)\to (0,0)}xy\cos\left(\frac{1}{x^{2}+y^{4}}\right)=0.$$ Therefore, $f$ is continuous at $(x,y)=(0,0)$.