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

$\underset{(x,y,z)\to (1,3,\pi )}{lim}\mathrm{sin}\frac{xz}{2y}$

bobbie71G
2021-10-15
Answered

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

$\underset{(x,y,z)\to (1,3,\pi )}{lim}\mathrm{sin}\frac{xz}{2y}$

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gotovub

Answered 2021-10-16
Author has **98** answers

To evaluate the limit: $\underset{(x,y,z)\to (1,3,\pi )}{lim}\mathrm{sin}\frac{xz}{2y}$

Evaluating the above limit.

$\underset{(x,y,z)\to (1,3,\pi )}{lim}\mathrm{sin}\frac{xz}{2y}=\mathrm{sin}\left(\frac{1cdo\pi}{2\cdot 3}\right)$

$\mathrm{sin}\left(\frac{\pi}{6}\right)$

$=\frac{1}{2}$

Therefore, limit of the function is$\frac{1}{2}$ .

We know that a function is continuous at a point x=a if:

$\underset{x\to a}{lim}f\left(x\right)=f\left(a\right)$

$\mathrm{sin}\left(\frac{xz}{2y}\right){\mid}_{(1,3,\pi )}=\mathrm{sin}\left(\frac{1\cdot \pi}{2\cdot 3}\right)$

$=\mathrm{sin}\frac{\pi}{6}$

$=\frac{1}{2}$

We can find that limit of function exists and also$\underset{(x,y,z)\to (1,3,\pi )}{lim}\mathrm{sin}\left(\frac{xz}{2y}\right)=\mathrm{sin}\left(\frac{xz}{2y}\right){\mid}_{(1,3,\pi )}$

Therefore, given function is continuous at given point.

Evaluating the above limit.

Therefore, limit of the function is

We know that a function is continuous at a point x=a if:

We can find that limit of function exists and also

Therefore, given function is continuous at given point.

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