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?
Checking the continuity of function at $(0,0)$
continuitylimits
Best Answer
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 $(0,0)$.