Checking the continuity of function at $(0,0)$ Let $f:\mathbb R^2\rightarrow\mathbb R$ be defined by $f(x,y)=\frac{1-\cos (x+y)}{x^2+y^2} $ and $f(0,0)=\frac{1}{2}$ then check the continuity of $f$ at $(0,0)$.
If $f$ is continuous at $(0,0)$ then $\lim_{(x,y) \to (0,0)} f(x,y)$ must exist and it must be equal to $\frac{1}{2}$. I think that the limit does not exist, as the numerator of f is bounded by 2 and the denominator is tending to zero. But how to show it using $\varepsilon \delta$ argument?
 A: Suppose $f$ function is defined on some $E \subset \mathbb{R}^n$ and point $x_0$ is density point for $E$. One of form of definition continuity is following:
$$\forall (x^{(m)}) \subset E, \lim_{m \rightarrow \infty}x^{(m)}= x_0 \Rightarrow \lim_{m \rightarrow \infty} f(x^{(m)})= f(x_0)$$
So for discontinuity we have
$$\exists (x^{(m)}) \subset E, \lim_{m \rightarrow \infty}x^{(m)}= x_0 \land \lim_{m \rightarrow \infty} f(x^{(m)}) \ne f(x_0)$$
In our case $x_0=(0,0)$. Let's consider $x^{(m)} = \left(\frac{1}{m},-\frac{1}{m}\right)$. It's obvious that $\lim_{m \rightarrow \infty}x^{(m)}= x_0$. And then we have $$f\left(\frac{1}{m},-\frac{1}{m}\right)=\frac{1-\cos (\frac{1}{m}-\frac{1}{m})}{\frac{1}{m^2}+\frac{1}{m^2}}=0$$
So limit cannot be $ \frac{1}{2}$.
A: Note that for $x\ne -y$:
$$
f(x,y)={1-\cos(x+y)\over x^2+y^2}={1-\cos(x+y)\over (x+y)^2}\cdot{(x+y)^2\over x^2+y^2}
$$Since
$$
\lim_{x,y\to 0}
 {1-\cos(x+y)\over (x+y)^2}={1\over 2}$$and ${(x+y)^2\over x^2+y^2}$ has no limit for $(x,y)\to (0,0)$ [check out the $x=ky$ branches], the function $f(x,y)$ is discontinuous in $(0,0)$.
