Limit and Continuity Find the limit (if it exists) and discuss the continuity of the...

To evaluate the limit: ${\displaystyle \underset{(x,y,z)\to (1,3,\pi )}{lim}\sin \frac{xz}{2y}}$
Evaluating the above limit.
${\displaystyle \underset{(x,y,z)\to (1,3,\pi )}{lim}\sin \frac{xz}{2y}=\sin \left(\frac{1cdo\pi }{2\cdot 3}\right)}$
${\displaystyle \sin \left(\frac{\pi }{6}\right)}$
${\displaystyle =\frac{1}{2}}$
Therefore, limit of the function is ${\displaystyle \frac{1}{2}}$.
We know that a function is continuous at a point x=a if:
${\displaystyle \underset{x\to a}{lim}f\left(x\right)=f\left(a\right)}$
${\displaystyle \sin \left(\frac{xz}{2y}\right){\mid }_{(1,3,\pi )}=\sin \left(\frac{1\cdot \pi }{2\cdot 3}\right)}$
${\displaystyle =\sin \frac{\pi }{6}}$
${\displaystyle =\frac{1}{2}}$
We can find that limit of function exists and also ${\displaystyle \underset{(x,y,z)\to (1,3,\pi )}{lim}\sin \left(\frac{xz}{2y}\right)=\sin \left(\frac{xz}{2y}\right){\mid }_{(1,3,\pi )}}$
Therefore, given function is continuous at given point.