At what points is the given function continuous?
-$$f ( x , y , z ) = y z \cos \left( \frac { 1 } { x } \right)$$
A) All $( x , y , z )$ such that $0 \leq \frac { 1 } { x } \leq 2 \pi$
B) All $( x , y , z )$ such that $x \neq 0$
C) All $( x , y , z )$
D) All $( x , y , z )$ such that $x \neq \frac { 2 } { \pi }$

Question

Quizplus · Accepted Answer

Note that the function is a product of continuous functions, namely $y$, $z$, and $\cos \left( \frac { 1 } { x } \right)$. Thus, the only possible discontinuity would be at $x = 0$ since that would cause the denominator of $\frac { 1 } { x }$ to be undefined. However, it is clear that the function approaches 0 as $(x, y, z)$ approaches $(0, 0, 0)$, so the function is continuous everywhere except possibly at $(0, 0, 0)$. Since the domain of the function is not specified, we assume that it is defined for all $(x, y, z)$, except probably $(0,0,0)$. Thus, the only choice that must be false is choice A, since it specifies a restricted domain where the function is continuous. Choice C is true, but not the best answer, as it implies that the function is continuous at $(0,0,0)$, which requires further justification. Choice D is also true, but it excludes a single point, which does not affect the continuity of the function. Thus, the best choice is B, which is the weakest sufficient condition for continuity.

At What Points Is the Given Function Continuous? - f(x,y,z)=yzcos⁡(1x)f ( x , y , z ) = y z \cos \left( \frac { 1 } { x } \right)f(x,y,z)=yzcos(x1​)

At What Points Is the Given Function Continuous?
- $f ( x , y , z ) = y z \cos \left( \frac { 1 } { x } \right)$