Evaluate the limits in Exercises 33–40 if they exist

$\underset{(x,y)\to (1,2)}{\lim }\frac{x^2+y^2}{x^2-y^2}$

Consider the phrase $\underset{(x,y)\to (1,2)}{\lim }\frac{x^2+y^2}{x^2-y^2}$

The goal is to assess $\underset{(x,y)\to (1,2)}{\lim }\frac{x^2+y^2}{x^2-y^2}$if it exists.

Take the following assertion:

Consider a two-variable function $f(x,y)$that is continuous at all points on $R^2$.

The limit of the function $f(x,y)as(x,y)\to (x_0,y_0)$is therefore defined as

$\underset{(x,y)\to (x_0,y_0)}{\lim }f(x_0,y_0)$

Because $x^2+y^2$and $x^2-y^2$are polynomial functions of two variables, they are continuous for all points on $R^2$.

As a result, the rational function $\frac{x^2+y^2}{x^2-y^2}$is continuous at all positions where $\frac{x^2+y^2}{x^2-y^2}$is defined.

The rational function $\frac{x^2+y^2}{x^2-y^2}$is discontinuous where $x^2-y^2=0$, that is,

$$\begin{gathered} x^2=y^2 \\ x=\pm y \end{gathered}$$

Because $x=1andy=2$do not satisfy the equation, the rational function $\frac{x^2+y^2}{x^2-y^2}$ is continuous at $(1,2)$

As a result of the statement,

$\underset{(x,y)\to (1,2)}{\lim }\frac{x^2+y^2}{x^2-y^2}=\frac{1+2^2}{1-2^2}=-\frac{5}{3}$