Finding the limit with three variables Here is the original problem: 
$$\lim_{(x,y,z)\to (0,0,0)}{(\cos x-1)\sin(2y)(e^{3z}-1)\over x^2yz}$$
I was thinking about splitting up the limit like this:
$$\lim_{(x,y,z)\to (0,0,0)}{(\cos x-1)\over x^2}\cdot\lim_{(x,y,z)\to (0,0,0)}{\sin(2y)\over y}\cdot\lim_{(x,y,z)\to (0,0,0)}{e^{3z}-1\over z}$$
And then break it down some more:$$\lim_{x\to 0}{(\cos x-1)\over x^2}\cdot\lim_{y\to 0}{\sin(2y)\over y}\cdot\lim_{z\to 0}{e^{3z}-1\over z}$$
And then I want to use L'Hospital's rule. If someone could let me know if I'm heading in the right direction or not, that would be great. Thanks!

$$\lim_{(x,y,z)\to (0,0,0)}{(\cos x-1)\sin(2y)(e^{3z}-1)\over x^2yz}$$
I was thinking about splitting up the limit like this:
$$\lim_{(x,y,z)\to (0,0,0)}{(\cos x-1)\over x^2}\cdot\lim_{(x,y,z)\to (0,0,0)}{\sin(2y)\over y}\cdot\lim_{(x,y,z)\to (0,0,0)}{e^{3z}-1\over z}$$
And then break it down some more:$$\lim_{x\to 0}{(\cos x-1)\over x^2}\cdot\lim_{y\to 0}{\sin(2y)\over y}\cdot\lim_{z\to 0}{e^{3z}-1\over z}$$
LH rule to the first part gives you (-0.5) 
Second part ofcourse gives you 2 by multiplying dividing by 2 and cancelling sin 2y/2yThird part again use LH rule to get 3.
Multiply all three  parts to eventually get an  ANSWER -3
