Let $f ( x , y ) = \left\{ \begin{array} { c c }
\frac { \sin ( x - y ) } { x - y } & x \neq y \
1 & x = y
\end{array} \right.$ . Then f is

A)Continuous at (2,2)
B)Discontinuous at (2,2) because f is undefined at (2,2)
C)Discontinuous at (2,2) because the limit of f at (2,2) does not exist
D)Discontinuous at (2,2) because the limit of f at (2,2) is different from the function values at (2,2)
E)Discontinuous at (2,2) for a reason that is different from the ones above

Question

Let $f ( x , y ) = \left\{ \begin{array} { c c } 
\frac { \sin ( x - y ) } { x - y } & x \neq y \
1 & x = y
\end{array} \right.$ . Then f is

A)Continuous at (2,2)
B)Discontinuous at (2,2) because f is undefined at (2,2)
C)Discontinuous at (2,2) because the limit of f at (2,2) does not exist
D)Discontinuous at (2,2) because the limit of f at (2,2) is different from the function values at (2,2)
E)Discontinuous at (2,2) for a reason that is different from the ones above

Quizplus · Accepted Answer

The function $f(x, y)$ is defined to be 1 when $x = y$, including at the point (2, 2). For $x \neq y$, it is given by the expression $\frac{\sin(x - y)}{x - y}$, which is continuous for all $x \neq y$. The definition directly provides the value of the function as 1 at (2, 2), ensuring continuity at that point.

Let f(x,y)={sin⁡(x−y)x−yx≠y1x=yf ( x , y ) = \left\{ \begin{array} { c c } \frac { \sin ( x - y ) } { x - y } & x \neq y \\1 & x = y\end{array} \right.f(x,y)={x−ysin(x−y)​1​x=yx=y​

Let $f ( x , y ) = \left\{ \begin{array} { c c } \frac { \sin ( x - y ) } { x - y } & x \neq y \\1 & x = y\end{array} \right.$