Question

Determine the set of points at which the function is continuous
1) G(x,y)=tan^-1((x+y)^-2)
2) f(x,y,z)=√y÷(x^2-y^2+z^2)
3) f(x,y,z)=√z+y+z

4) f(x,y)={x^2y^3÷(2x^2+y^2 if (x,y)≠(0,0)
1 if (x,y)=(0,0)

5) f(x,y)={ xy÷(x^2+xy+y^2 if (x,y)≠(0,0)
0 if (x,y)=(0,0)

Answer 1

1. Now $tan^{-1}$ is continuous function. Hence given function is continuous except when $x+y=0$ in which case the function is not defined. Hence the function is continuous in its domain of definition i.e. $x+y\neq 0.$
2. In this function also, being a rational function is continuous every where in its domain of definition, namely, $y\geq 0, x^2-y^2+z^2\neq 0.$
3. This function is continuous at every point in its domain of definition, i.e. $z\geq 0.$
4. Let $x\neq 0.$ Let $y=mx.$ Then $f(x,mx)=\frac{m^3x^3}{2+m^2}$ This function is always continuous. Now if $x=0,$ then $f(x,y)=0$ if $y\neq 0.$ Now for any sequence $(x_n,y_n)\rightarrow (0,y), f(x_n,y_n)\rightarrow 0.$ Hence continuous. Now only point left is $(0,0).$ Now if we take, $x=y,$ we get, $f(x,y)=\frac{x^3}{3}.$ So $lim _{x\rightarrow 0} f(x,x)=0.$ But $f(0,0)=1.$ Hence function is continuous everywhere except at (0,0).
5. As in 4, $f(x,mx)=\frac{m}{1+m+m^2}$. Hence continuous for $x\neq 0.$ Now if $x=0,$ then the function takes value 0. So for $(0,y), y\neq 0,$ $f(x,y)=0.$ Also for any sequence $(x_n,y_n)\rightarrow (0,y),$ $f(x_n,y_n)$$\rightarrow 0$ since numerator tends to 0, but denominator is non-zero. Hence continuous at all points except may be at (0,0). Now taking $y=x=1/n,$ we see that $f(x,y)=\frac{1}{3}$ . Hence this doesn't converge to $f(0,0)=0$ and hence not continuous at (0,0).