<s>[INST] Prove that limit $\lim \limits_{(x,y,z) \to (0,0,0)} \frac{xyz\tan(2x+y-z)}{(2x+y)^2-z^2}$ doesn't exist Show that the Limit doesn't exist
$\lim \limits_{(x,y,z) \to (0,0,0)} \frac{xyz\tan(2x+y-z)}{(2x+y)^2-z^2}$
I try to replace all of x, y, and z with what they are actually approaching. Please help me to prove that the Limit doesn't exist
 [/INST] As suggested we have
$$\frac{\tan(2x+y-z)}{2x+y-z}\cdot\frac{xyz(2x+y-z)}{(2x+y)^2-z^2}$$
with $\frac{\tan(2x+y-z)}{2x+y-z}\to 1$ and
$$\frac{xyz(2x+y-z)}{(2x+y)^2-z^2}=\frac{xyz(2x+y-z)}{(2x+y-z)(2x+y+z)}=\frac{xyz}{2x+y+z}$$
which tends to zero for $(x,y,z)=(0,t,t)$ with $t \to 0$ but for $(x,y,z)=(t,-t,-t+t^3)$ we obtain
$$\frac{xyz}{2x+y+z}=\frac{t^3-t^5}{t^3}=1-t^2 \to 1$$
therefore the limit doesn't exist.
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