Let $$ f(x,y)= \begin{cases} \frac{1-\cos(x+y)}{x^2+y^2}, &(x,y)\neq 0\\ \frac{1}{2}, & (x,y)= 0 \end{cases} $$ and $$ g(x,y)= \begin{cases} \frac{1-\cos(x+y)}{\left({x+y}\right)^2}, & x+y\neq 0\\ \frac{1}{2}, & x+y=0 \end{cases} $$
$f$ is continuous at $(0,0)$
$f$ is continuous everywhere except at $(0,0)$
$g$ is continuous at $(0,0)$
$g$ is continuous everywhere
My attempt I tried the sequential criterion using the sequence I took the sequence $(x_n,y_n)=(1/n,-1/n)$. Then, $f(x_n,y_n)=\frac{0}{1/n^2}$ which converges to $0$. So, $f(x,y)$ is not continuous at $(x,y)=(0,0)$. So, (1) and (2) are right.
I tried counter example for the sequence contradict the continuity. I couldn't find. I tried to prove, for given $\varepsilon>0$, $|f(x,y)-f(0,0)|=\left|\frac{1-\cos(x+y)}{x^2+y^2}-\frac{1}{2}\right| = \left|\frac{2-2\cos(x+y)-x^2-y^2}{2(x^2+y^2)}\right|$. How do I proceed further and find the desired neighbourhood?
Nice choice of sequence for the first part.
For $g$, note that $g$ is clearly continuous on the plane with $x=-y$ deleted. This deleted line includes the origin also.
So you only need to consider the image of the set $\{x \approx -y\}$ under $g$, in other words, $x = -y + h$ with $0 < |h|\ll 1$. Plug in $x = -y + h$ into $g$ and then take limits :)