Q.Show that $f(x,y)=\dfrac{x^2-y^2}{y^2+1}$ is continuous at every $(x,y)\in\mathbb R^2$.

A function of one variable $y=f(x)$ traces a curve in the $xy$-plane. A function of two variables $F(x,y)$ is visualized by graphing $z=F(x,y)$, which traces a surface in $xyz$-space: the point $(x,y,F(x,y))$ sits $F(x,y)$ units above (or below) the point $(x,y)$ in the $xy$-plane.

Fixing one variable slices the surface with a plane and produces a curve: for $g(x,y)=30-x^2-y^2$, holding $y=3$ gives $g(x,3)=21-x^2$ (a parabola, the intersection of the surface with the plane $y=3$), and holding $x=2$ gives $g(2,y)=26-y^2$. The surface $z=30-x^2-y^2$ itself is called a paraboloid. This is the natural generalization from one to several variables: profit as a function of the units of two products, or volume as a function of length, width and height, are genuinely functions of more than one variable and cannot be reduced to a single-variable picture.

Neighbourhoods in $\mathbb R^2$. To define limits and continuity for $F(x,y)$, replace the one-variable interval-neighbourhood $(x_0-\delta,x_0+\delta)$ by an open disc:

— the set of points strictly within distance $r$ of $(u,v)$. Removing the centre gives a deleted neighbourhood.

Definition (Limit of a Function of Two Variables). $F$ has limit $L$ at $(u,v)$, written $\displaystyle\lim_{(x,y)\to(u,v)}F(x,y)=L$, if for every neighbourhood $(L-\varepsilon,L+\varepsilon)$, $\varepsilon>0$, of $L$ there exists a $\delta$-neighbourhood $B_\delta((u,v))$ of $(u,v)$ such that $(x,y)\in B_\delta((u,v))\setminus\{(u,v)\}\ \Rightarrow\ F(x,y)\in(L-\varepsilon,L+\varepsilon)$.

Definition (Continuity). $F$ is continuous at $(u,v)$ if (1) $F(u,v)$ is defined, (2) $\displaystyle\lim_{(x,y)\to(u,v)}F(x,y)$ exists, and (3) that limit equals $F(u,v)$ — exactly the same three-part test as one variable, now over $\mathbb R^2$. All the standard limit theorems (limits of sums, products, quotients, composition with a continuous function) carry over unchanged from one variable to several. …