bobbie71G

2021-10-15

Limit and Continuity Find the limit (if it exists) and discuss the continuity of the function.
$\underset{\left(x,y,z\right)\to \left(1,3,\pi \right)}{lim}\mathrm{sin}\frac{xz}{2y}$

gotovub

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

Jeffrey Jordon