Determine continuity and differentiability Could you tell me how to determine continuity and differentiability of functions with two or more variables?
I know how to do it for fairly simple functions, such as $f(x,y)=\sqrt{x^k + y^k}$, $f(x,y)=\sin (x^k + y^k)$ and for polynomials continuity is obvious.
We check if the function and its partial derivatives are continuous.
For functions for which it is evident that they are a composition of continuous functions, it is easy to establish continuity.
I have problems determining continuity of more ambitious functions, say $f(x,y)  = \sqrt{xy}(\sqrt{x^2+y^2})^{-1}$ for $(x,y)$ in the domain of this function and $f(x,y) = 0$  elsewhere.
I've also seen functions of this type $(|x|^ky) \cdot (x^2 + x^4)^{-1} $, and I also don't know how to deal with them.
How to calculate $\lim _{(x,y)\rightarrow (0,0)}f(x,y)$ here?
I would appreciate it if you explained to me how to solve such problems on an example of the functions I've mentioned above. 
Thank you.
 A: When the function is not continuous, you can prove it by going to the origin through two special (may not work for any two) curves or sequences so that the function approaches two different values. For example, consider the function
$$
f(x, y) = \frac{xy}{x^2 + y^2}
$$
for $(x, y) \neq (0, 0)$ and $f(0, 0) = 0$. We can go along the curve $y = x$ to get
$$
\lim_{x \to 0} f(x, x) = \lim_{x \to 0} \frac{x \cdot x}{x^2 + x^2} = \frac{1}{2}
$$
whereas going along $y = -x$, we get
$$
\lim_{x \to 0} f(x, -x) = \lim_{x \to 0} \frac{x \cdot (-x)}{x^2 + (-x)^2} = -\frac{1}{2}.
$$
So $f$ is clearly not continuous at $(0, 0)$.
To prove continuity, one way is to bound the function by something else which you can show goes to $0$. For example, consider the function
$$
f(x, y) = \frac{|xy|}{\sqrt{x^2 + y^2}}
$$
for $(x, y) \neq (0, 0)$ and $f(0, 0) = 0$. Then, $0 \leq (|x| - |y|)^2 \implies 2|xy| \leq x^2 + y^2$. Thus for $(x, y) \neq (0, 0)$,
$$
0 \leq \frac{|xy|}{\sqrt{x^2 + y^2}} \leq \frac{x^2 + y^2}{2\sqrt{x^2 + y^2}} = \frac{1}{2}\sqrt{x^2 + y^2} \\
\implies 0 \leq \lim_{(x, y) \to (0, 0)} f(x, y) = \lim_{(x, y) \to (0, 0)} \frac{|xy|}{\sqrt{x^2 + y^2}} \leq \lim_{(x, y) \to (0, 0)} \frac{1}{2}\sqrt{x^2 + y^2} = 0
$$
and hence the limit is $\lim_{(x, y) \to (0, 0)} f(x, y) = 0$ which shows $f$ is continuous.
