Continuity of $\sqrt{1-x^2-y^2}$ I saw this question on an interim exam; Let $f: \mathbb{R^2} \rightarrow \mathbb{R}$ be given by 

$f(x,y)= \left\{ 
\begin{array}{}
             \sqrt{1-x^2-y^2}  &amp;x^2+y^2 \leq 1, \ y \geq 0 \\
              0 &amp;\text{Otherwise}
\end{array}
\right.$

Now, the question was: Is $f$ continuous in $(0,0)$, $(1,0)$, $(1,1)$. Using the fact that $f$ is continuous in $a=(a_1,a_2)$ iff $\lim_{(x,y)\rightarrow (a_1,a_2)}f(x,y) = f(a_1,a_2)$, I thought that $f$ is continuous in $(0,0)$ and $(1,0)$, although I am not sure about $(1,1)$. 

Could someone verify that my assumption about the first two points is correct, and help me verifying the last point? Thanks in advance!
Let $(x_n) $ and $(y_n) $ such that

$x_n\to 1$ and $y_n\to 1$.

then for $n $ great enough,

$x_n&gt;\frac{\sqrt {2}}{2}$ and $y_n&gt;\frac {\sqrt {2}}{2}$.
thus $x_n^2+y_n^2&gt;1$ and
$$f (x_n,y_n)=0=\lim_{n\to+\infty}f (x_n,y_n)=f (1,1) $$

$f $ is continuous at $(1,1) $.


  for $(0,0) $.


$$\lim_{+\infty}f (0,1/n)=1$$
$$ \lim_{+\infty}f (0,-1/n)=0$$
$f $ not continuous at $(0,0) $.
Let $(x_n) $ and $(y_n) $ such that

$x_n\to 1$ and $y_n\to 1$.

then for $n $ great enough,

$x_n&gt;\frac{\sqrt {2}}{2}$ and $y_n&gt;\frac {\sqrt {2}}{2}$.
thus $x_n^2+y_n^2&gt;1$ and
$$f (x_n,y_n)=0=\lim_{n\to+\infty}f (x_n,y_n)=f (1,1) $$

$f $ is continuous at $(1,1) $.


  for $(0,0) $.


$$\lim_{+\infty}f (0,1/n)=1$$
$$ \lim_{+\infty}f (0,-1/n)=0$$
$f $ not continuous at $(0,0) $.