multivariable limit that DNE or is equal to 0

Question

My classmates and I were told to evalulate the following limit; however, we have 2 different answers. My answer was "does not exist" since if we plugged in $\frac{1}{2}$ for $z$, we'd get $\sin(\frac{1}{0})$. Others claim that the limit goes to 0. Which would be the correct answer? $$\lim_{(x,y,z)\to (1/2,1/2,1/2)}{x\cos(\pi y)\sin\left(\frac{1}{2z-1}\right)}$$

Answer 1

Your friends are right, though I completely understand your answer. The technical definition of a limit of a multivariable function $f(x,y,z)$ is $$ \lim_{(x,y,z)\to(x_0,y_0,z_0)}f(x,y,z)=L $$ if for every path in the domain of $f$ that converges to $(x_0,y_0,z_0)$ results in $f$ converging to $L$. In this case, $$ 0\leq \Big | x\cos(y\pi)\sin\left(\frac{1}{2z-1}\right)\Big |\leq |x|\,|\cos(y\pi)|. $$ The right hand side converges to $0$ for any path since both functions are continuous. Then apply squeeze theorem.